B. Baby Bites水题直接模拟即可
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define pi acos(-1.0)
13 #define eps 1e-9
14 #define fi first
15 #define se second
16 #define rtl rt<<1
17 #define rtr rt<<1|1
18 #define bug printf("******\n")
19 #define mem(a,b) memset(a,b,sizeof(a))
20 #define name2str(x) #x
21 #define fuck(x) cout<<#x" = "<<x<<endl
22 #define f(a) a*a
23 #define sf(n) scanf("%d", &n)
24 #define sff(a,b) scanf("%d %d", &a, &b)
25 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
26 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
27 #define pf printf
28 #define FRE(i,a,b) for(i = a; i <= b; i++)
29 #define FREE(i,a,b) for(i = a; i >= b; i--)
30 #define FRL(i,a,b) for(i = a; i < b; i++)+
31 #define FRLL(i,a,b) for(i = a; i > b; i--)
32 #define FIN freopen("data.txt","r",stdin)
33 #define gcd(a,b) __gcd(a,b)
34 #define lowbit(x) x&-x
35 #define rep(i,a,b) for(int i=a;i<b;++i)
36 #define per(i,a,b) for(int i=a-1;i>=b;--i)
37 using namespace std;
38 typedef long long LL;
39 typedef unsigned long long ULL;
40 const int INF = 0x3f3f3f3f;
41 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
42 const int maxn = 1e5 + 10;
43 int n;
44 int main() {
45 sf ( n );
46 char op[10];
47 int flag = 1;
48 for ( int i = 1 ; i <= n ; i++ ) {
49 scanf ( "%s", op );
50 if ( op[0] >= ‘0‘ && op[0] <= ‘9‘ ) {
51 int temp = 0;
52 int len = strlen ( op );
53 for ( int j = 0 ; j < len ; j++ ) temp = temp * 10 + op[j] - ‘0‘;
54 if ( temp != i ) flag = 0;
55 }
56 }
57 if ( flag ) printf ( "makes sense\n" );
58 else printf ( "something is fishy\n" );
59 return 0;
60 }
C. Code Cleanups（难点）在于读题 读题读到自闭
题意： 给你一个数组a[ ] 表示a[ i ]产生一个垃圾，之后每天肮脏度增加1 ，肮脏度到达20时必须清理，问最少清理天数（注意每年的结束要将肮脏度变为0）
直接模拟即可
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define pi acos(-1.0)
13 #define eps 1e-9
14 #define fi first
15 #define se second
16 #define rtl rt<<1
17 #define rtr rt<<1|1
18 #define bug printf("******\n")
19 #define mem(a,b) memset(a,b,sizeof(a))
20 #define name2str(x) #x
21 #define fuck(x) cout<<#x" = "<<x<<endl
22 #define f(a) a*a
23 #define sf(n) scanf("%d", &n)
24 #define sff(a,b) scanf("%d %d", &a, &b)
25 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
26 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
27 #define pf printf
28 #define FRE(i,a,b) for(i = a; i <= b; i++)
29 #define FREE(i,a,b) for(i = a; i >= b; i--)
30 #define FRL(i,a,b) for(i = a; i < b; i++)+
31 #define FRLL(i,a,b) for(i = a; i > b; i--)
32 #define FIN freopen("data.txt","r",stdin)
33 #define gcd(a,b) __gcd(a,b)
34 #define lowbit(x) x&-x
35 #define rep(i,a,b) for(int i=a;i<b;++i)
36 #define per(i,a,b) for(int i=a-1;i>=b;--i)
37 using namespace std;
38 typedef long long LL;
39 typedef unsigned long long ULL;
40 const int INF = 0x3f3f3f3f;
41 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
42 const int maxn = 1e5 + 10;
43 int n, vis[maxn];
44 int main() {
45 sf ( n );
46 for ( int i = 0, x ; i < n ; i++ ) sf ( x ), vis[x] = 1;
47 int sum = 0, cnt = 0, ans = 0;
48 for ( int i = 1 ; i <= 365 ; i++ ) {
49 sum += cnt;
50 if ( sum >= 20 ) sum = 0, ans++, cnt = 0;
51 if ( vis[i] ) cnt++;
52 }
53 if ( cnt || sum ) ans++;
54 printf ( "%d\n", ans );
55 return 0;
56 }
E. Explosion Exploit 记忆化搜索
题意：你有n个士兵，敌方有m个士兵，每一个士兵都有一定的血量（最大为6），如果血量归零，则证明该士兵死亡。现在有d点伤害，每一点伤害都会以等概率分配给任意一个人。现在问你，敌方的m个士兵全都阵亡的概率。
题解：总共只有12个士兵用一个long long 就可以存好状态了 当status<1000000时表示敌军已经死完了
这个概率是倒推过来的
Dfs(status,num)表示状态为status，剩下num点伤害，对方阵亡的概率
ans += ( double ) mp[i][j] / ( double ) sum * res;//表示转移到下一个状态的概率
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <vector>
12 #include <bits/stdc++.h>
13 #define pi acos(-1.0)
14 #define eps 1e-9
15 #define fi first
16 #define se second
17 #define rtl rt<<1
18 #define rtr rt<<1|1
19 #define bug printf("******\n")
20 #define mem(a,b) memset(a,b,sizeof(a))
21 #define name2str(x) #x
22 #define fuck(x) cout<<#x" = "<<x<<endl
23 #define f(a) a*a
24 #define sf(n) scanf("%d", &n)
25 #define sff(a,b) scanf("%d %d", &a, &b)
26 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
27 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
28 #define pf printf
29 #define FRE(i,a,b) for(i = a; i <= b; i++)
30 #define FREE(i,a,b) for(i = a; i >= b; i--)
31 #define FRL(i,a,b) for(i = a; i < b; i++)+
32 #define FRLL(i,a,b) for(i = a; i > b; i--)
33 #define FIN freopen("data.txt","r",stdin)
34 #define gcd(a,b) __gcd(a,b)
35 #define lowbit(x) x&-x
36 #define rep(i,a,b) for(int i=a;i<b;++i)
37 #define per(i,a,b) for(int i=a-1;i>=b;--i)
38 using namespace std;
39 typedef long long LL;
40 typedef unsigned long long ULL;
41 const int INF = 0x3f3f3f3f;
42 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
43 const int maxn = 2505;
44 const int mod = 1e9 + 7;
45 int n, m, d, mp[2][10];
46 unordered_map<LL, double>dp;
47 LL getstatus() {
48 LL temp = 0;
49 for ( int i = 1 ; i <= 6 ; i++ ) temp = temp * 10 + mp[1][i];
50 for ( int i = 1 ; i <= 6 ; i++ ) temp = temp * 10 + mp[0][i];
51 return temp;
52 }
53 double dfs ( LL status, int num ) {
54 if ( status < 1000000 ) return 1;
55 if ( num == 0 ) return 0;
56 if (dp.count(status)) return dp[status];
57 double ans = 0;
58 int sum = 0;
59 for ( int i = 0 ; i < 2 ; i++ )
60 for ( int j = 1 ; j <= 6 ; j++ ) sum += mp[i][j];
61 for ( int i = 0 ; i < 2 ; i++ ) {
62 for ( int j = 1 ; j <= 6 ; j++ ) {
63 if ( !mp[i][j] ) continue;
64 mp[i][j]--, mp[i][j - 1]++;
65 LL sta=getstatus();
66 double res = dfs ( sta, num - 1 );
67 dp[sta]=res;
68 mp[i][j - 1]--, mp[i][j]++;
69 ans += ( double ) mp[i][j] / ( double ) sum * res;
70 }
71 }
72 return ans;
73 }
74
75 int main() {
76 sfff ( n, m, d );
77 for ( int i = 1, x ; i <= n ; i++ ) {
78 sf ( x );
79 mp[0][x]++;
80 }
81 for ( int i = 1, x; i <= m; i++ ) {
82 sf ( x );
83 mp[1][x]++;
84 }
85 double ans = dfs ( getstatus(), d );
86 printf ( "%.8f\n", ans );
87 return 0 ;
88 }
H. House Lawn
有一个 l 平方米的草坪，给你 m 台割草机。选择其中一台，在一周（10080分钟）的时间内割完草坪。每台割草机输入格式为：“名字,价格,割草效率,工作时间,充电时间” ；
要求选出价格最低的满足条件的一台，如果有多台满足条件且价格相同，按照输入顺序输出割草机名字。
一开始我以为只要一周内可以完成一次就好了，然后拿int写的，后面仔细看题要保证T周也必须清理T次，用int写的话你这周可以清理完，但是下周不一定。所有这题的就相当于充t分钟电能给割草（t/充电时间）*割草时间 所以用double写这题
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define pi acos(-1.0)
13 #define eps 1e-9
14 #define fi first
15 #define se second
16 #define rtl rt<<1
17 #define rtr rt<<1|1
18 #define bug printf("******\n")
19 #define mem(a,b) memset(a,b,sizeof(a))
20 #define name2str(x) #x
21 #define fuck(x) cout<<#x" = "<<x<<endl
22 #define f(a) a*a
23 #define sf(n) scanf("%d", &n)
24 #define sff(a,b) scanf("%d %d", &a, &b)
25 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
26 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
27 #define pf printf
28 #define FRE(i,a,b) for(i = a; i <= b; i++)
29 #define FREE(i,a,b) for(i = a; i >= b; i--)
30 #define FRL(i,a,b) for(i = a; i < b; i++)+
31 #define FRLL(i,a,b) for(i = a; i > b; i--)
32 #define FIN freopen("data.txt","r",stdin)
33 #define gcd(a,b) __gcd(a,b)
34 #define lowbit(x) x&-x
35 #define rep(i,a,b) for(int i=a;i<b;++i)
36 #define per(i,a,b) for(int i=a-1;i>=b;--i)
37 using namespace std;
38 typedef long long LL;
39 typedef unsigned long long ULL;
40 const int INF = 0x3f3f3f3f;
41 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
42 const int maxn = 1e2 + 5;
43 string s[maxn], s1;
44 LL a[maxn], l, n, cnt[10], vis[maxn];
45 int main() {
46 // FIN;
47 cin >> l >> n;
48 getchar();
49 int ans = INF;
50 for ( int i = 0; i < n ; i++ ) {
51 getline ( cin, s1 );
52 int flag = 0, temp;
53 for ( int j = 0 ; j < s1.size() ; j++ ) {
54 if ( s1[j] == ‘,‘ ) flag++, cnt[flag] = temp, temp = 0;
55 else if ( !flag ) s[i] += s1[j];
56 else temp = temp * 10 + s1[j] - ‘0‘;
57 }
58 cnt[++flag] = temp;
59 a[i] = cnt[2];
60 // int num = 10080 / ( cnt[4] + cnt[5] ), res = 10080 % ( cnt[4] + cnt[5] );
61 // for (int i=2 ;i<=5 ;i++) printf("%d%c",cnt[i],i==5?‘\n‘:‘ ‘);
62 // printf ( "num = %d res = %d\n", num, res );
63 // LL sum = 0;
64 // if ( res >= cnt[4] ) sum = 1LL * ( num + 1 ) * 1LL * cnt[4] * 1LL * cnt[3];
65 // if ( res < cnt[4] ) sum = 1LL * num * 1LL * cnt[4] * 1LL * cnt[3] + 1LL * res * 1LL * cnt[3];
66 if ( ( double ) ( ( double ) 10080 / ( ( double ) ( cnt[4] + cnt[5] ) ) * ( double ) cnt[3] * ( double ) cnt[4] ) >= 1.0 * l && ans >= a[i] ) ans = cnt[2],vis[i]=1;
67 }
68 if ( ans == INF ) printf ( "no such mower\n" );
69 for ( int i = 0 ; i < n ; i++ ) if ( ans == a[i] && vis[i]) cout << s[i] << endl;
70 return 0;
71 }
Intergalactic Bidding
有n个人的名字和值，每个人的值相差两倍以上，问是否有一种方案使部分人的值加起来为S
大数直接模拟就好了（相差2倍以上，只有唯一解）
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define pi acos(-1.0)
13 #define eps 1e-9
14 #define fi first
15 #define se second
16 #define rtl rt<<1
17 #define rtr rt<<1|1
18 #define bug printf("******\n")
19 #define mem(a,b) memset(a,b,sizeof(a))
20 #define name2str(x) #x
21 #define fuck(x) cout<<#x" = "<<x<<endl
22 #define f(a) a*a
23 #define sf(n) scanf("%d", &n)
24 #define sff(a,b) scanf("%d %d", &a, &b)
25 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
26 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
27 #define pf printf
28 #define FRE(i,a,b) for(i = a; i <= b; i++)
29 #define FREE(i,a,b) for(i = a; i >= b; i--)
30 #define FRL(i,a,b) for(i = a; i < b; i++)+
31 #define FRLL(i,a,b) for(i = a; i > b; i--)
32 #define FIN freopen("data.txt","r",stdin)
33 #define gcd(a,b) __gcd(a,b)
34 #define lowbit(x) x&-x
35 #define rep(i,a,b) for(int i=a;i<b;++i)
36 #define per(i,a,b) for(int i=a-1;i>=b;--i)
37 using namespace std;
38 typedef long long LL;
39 typedef unsigned long long ULL;
40 const int INF = 0x3f3f3f3f;
41 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
42 const int MAXL = 2500;
43 const int MAXN = 9999;
44 const int DLEN = 4;
45 class Big {
46 public:
47 int a[MAXL], len;
48 Big ( const int b = 0 ) {
49 int c, d = b;
50 len = 0;
51 memset ( a, 0, sizeof ( a ) );
52 while ( d > MAXN ) {
53 c = d - ( d / ( MAXN + 1 ) ) * ( MAXN + 1 );
54 d = d / ( MAXN + 1 );
55 a[len++] = c;
56 }
57 a[len++] = d;
58 }
59 Big ( const char *s ) {
60 int t, k, index, L;
61 memset ( a, 0, sizeof ( a ) );
62 L = strlen ( s );
63 len = L / DLEN;
64 if ( L % DLEN ) len++;
65 index = 0;
66 for ( int i = L - 1; i >= 0; i -= DLEN ) {
67 t = 0;
68 k = i - DLEN + 1;
69 if ( k < 0 ) k = 0;
70 for ( int j = k; j <= i; j++ ) t = t * 10 + s[j] - ‘0‘;
71 a[index++] = t;
72 }
73 }
74 Big operator/ ( const LL &b ) const {
75 Big ret;
76 LL down = 0;
77 for ( int i = len - 1; i >= 0; i-- ) {
78 ret.a[i] = ( a[i] + down * ( MAXN + 1 ) ) / b;
79 down = a[i] + down * ( MAXN + 1 ) - ret.a[i] * b;
80 }
81 ret.len = len;
82 while ( ret.a[ret.len - 1] == 0 && ret.len > 1 ) ret.len--;
83 return ret;
84 }
85 bool operator> ( const Big &T ) const {
86 int ln;
87 if ( len > T.len ) return true;
88 else if ( len == T.len ) {
89 ln = len - 1;
90 while ( a[ln] == T.a[ln] && ln >= 0 ) ln--;
91 if ( ln >= 0 && a[ln] > T.a[ln] ) return true;
92 else return false;
93 } else return false;
94 }
95 Big operator+ ( const Big &T ) const {
96 Big t ( *this );
97 int big = T.len > len ? T.len : len;
98 for ( int i = 0; i < big; i++ ) {
99 t.a[i] += T.a[i];
100 if ( t.a[i] > MAXN ) {
101 t.a[i + 1]++;
102 t.a[i] -= MAXN + 1;
103 }
104 }
105 if ( t.a[big] != 0 ) t.len = big + 1;
106 else t.len = big;
107 return t;
108 }
109 Big operator- ( const Big &T ) const {
110 int big;
111 bool flag;
112 Big t1, t2;
113 if ( *this > T ) {
114 t1 = *this;
115 t2 = T;
116 flag = 0;
117 } else {
118 t1 = T;
119 t2 = *this;
120 flag = 1;
121 }
122 big = t1.len;
123 for ( int i = 0; i < big; i++ ) {
124 if ( t1.a[i] < t2.a[i] ) {
125 int j = i + 1;
126 while ( t1.a[j] == 0 ) j++;
127 t1.a[j--]--;
128 while ( j > i ) t1.a[j--] += MAXN;
129 t1.a[i] += MAXN + 1 - t2.a[i];
130 } else t1.a[i] -= t2.a[i];
131 }
132 t1.len = big;
133 while ( t1.a[t1.len - 1] == 0 && t1.len > 1 ) {
134 t1.len--;
135 big--;
136 }
137 if ( flag ) t1.a[big - 1] = 0 - t1.a[big - 1];
138 return t1;
139 }
140 LL operator% ( const int &b ) const {
141 LL d = 0;
142 for ( int i = len - 1; i >= 0; i-- ) d = ( ( d * ( MAXN + 1 ) ) % b + a[i] ) % b;
143 return d;
144 }
145 Big operator* ( const Big &T ) const {
146 Big ret;
147 int i, j, up, temp, temp1;
148 for ( i = 0; i < len; i++ ) {
149 up = 0;
150 for ( j = 0; j < T.len; j++ ) {
151 temp = a[i] * T.a[j] + ret.a[i + j] + up;
152 if ( temp > MAXN ) {
153 temp1 = temp - temp / ( MAXN + 1 ) * ( MAXN + 1 );
154 up = temp / ( MAXN + 1 );
155 ret.a[i + j] = temp1;
156 } else {
157 up = 0;
158 ret.a[i + j] = temp;
159 }
160 }
161 if ( up != 0 ) ret.a[i + j] = up;
162 }
163 ret.len = i + j;
164 while ( ret.a[ret.len - 1] == 0 && ret.len > 1 ) ret.len--;
165 return ret;
166 }
167 void print() {
168 printf ( "%d", a[len - 1] );
169 for ( int i = len - 2; i >= 0; i-- ) printf ( "%04d", a[i] );
170 }
171 };
172 int n;
173 struct node {
174 char name[101];
175 char num[2010];
176 } qu[MAXL];
177 int cmp ( node s1, node s2 ) {
178 int len1 = strlen ( s1.num ), len2 = strlen ( s2.num );
179 if ( len2 > len1 ) return 1;
180 else if ( len2 < len1 ) return 0;
181 else {
182 for ( int i = 0 ; i < len1 ; i++ )
183 if ( s2.num[i] < s1.num[i] ) return 0;
184 else if ( s2.num[i] > s1.num[i] ) return 1;
185 }
186 }
187 char sum[2010];
188 Big s, a[MAXL];
189 vector<int>ans;
190 int main() {
191 scanf ( "%d%s", &n, sum );
192 s = sum;
193 for ( int i = 0 ; i < n ; i++ ) scanf ( "%s%s", qu[i].name, qu[i].num ) ;
194 sort ( qu, qu + n, cmp );
195 //for ( int i = 0 ; i < n ; i++ ) printf ( "%s %s\n", qu[i].name, qu[i].num );
196 for ( int i = 0 ; i < n ; i++ ) a[i] = qu[i].num;
197 int flag = 0;
198 for ( int i = n - 1 ; i >= 0 ; i-- ) {
199 // a[i].print(), printf ( " " ), s.print(), printf ( "\n" );
200 if ( a[i] > s ) continue;
201 else if ( s > a[i] ) {
202 s = s - a[i];
203 ans.push_back ( i );
204 } else {
205 ans.push_back ( i );
206 flag = 1;
207 break;
208 }
209 }
210 if ( !flag ) printf ( "0\n" );
211 else {
212 printf ( "%d\n", ans.size() );
213 for ( int i = 0 ; i < ans.size() ; i++ )
214 printf ( "%s\n", qu[ans[i]].name );
215 }
216 return 0;
217 }
J. Jumbled String
有a个00子顺序串，b个01，c个10，d个11，构造01串
注意一点当全部等于0的时候要输出0或者1
还有a==0 ，0的个数有0或者1个 ，d==0 同理
我的构造是111100001111 然后再0之中插入一个1构造01和10
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define pi acos(-1.0)
13 #define eps 1e-9
14 #define fi first
15 #define se second
16 #define rtl rt<<1
17 #define rtr rt<<1|1
18 #define bug printf("******\n")
19 #define mem(a,b) memset(a,b,sizeof(a))
20 #define name2str(x) #x
21 #define fuck(x) cout<<#x" = "<<x<<endl
22 #define f(a) a*a
23 #define sf(n) scanf("%d", &n)
24 #define sff(a,b) scanf("%d %d", &a, &b)
25 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
26 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
27 #define pf printf
28 #define FRE(i,a,b) for(i = a; i <= b; i++)
29 #define FREE(i,a,b) for(i = a; i >= b; i--)
30 #define FRL(i,a,b) for(i = a; i < b; i++)+
31 #define FRLL(i,a,b) for(i = a; i > b; i--)
32 #define FIN freopen("data.txt","r",stdin)
33 #define gcd(a,b) __gcd(a,b)
34 #define lowbit(x) x&-x
35 #define rep(i,a,b) for(int i=a;i<b;++i)
36 #define per(i,a,b) for(int i=a-1;i>=b;--i)
37 using namespace std;
38 typedef long long LL;
39 typedef unsigned long long ULL;
40 const int INF = 0x3f3f3f3f;
41 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
42 const int maxn = 1e2 + 5;
43 map<LL, int>mp;
44 LL a, b, c, d;
45 int main() {
46 for ( int i = 2 ; i <= 100001 ; i++ ) mp[ ( i - 1 ) *i / 2] = i;
47 scanf ( "%lld%lld%lld%lld", &a, &b, &c, &d );
48 int num0 = mp[a], num1 = mp[d];
49 if ( a == 0 && b == 0 && c == 0 && d == 0 ) return 0 * printf ( "1\n" );
50 if ( ( num0 == 0 && a != 0 ) || ( num1 == 0 && d != 0 ) ) return 0 * printf ( "impossible\n" );
51 if ( num0 == 0 && ( b || c ) ) num0 = 1;
52 if ( num1 == 0 && ( b || c ) ) num1 = 1;
53 if ( num0 * num1 != b + c ) return 0 * printf ( "impossible\n" );
54 if ( num0 == 0 && b == 0 && c == 0 ) {
55 for ( int i = 0 ; i < num1 ; i++ ) printf ( "1" );
56 printf ( "\n" );
57 return 0;
58 }
59 if ( num1 == 0 && b == 0 && c == 0 ) {
60 for ( int i = 0 ; i < num0 ; i++ ) printf ( "0" );
61 printf ( "\n" );
62 return 0;
63 }
64 //printf ( "num0 = %d num1 = %d\n", num0, num1 );
65 int cnt = c / num0, res = c % num0;
66 for ( int i = 0 ; i < cnt ; i++ ) printf ( "1" );
67 for ( int i = 0 ; i < num0 - max ( res - ( res == 0 ), 0 ) ; i++ ) printf ( "0" );
68 if ( res != 0 ) printf ( "1" ), num1--;
69 for ( int i = 0; i < max ( res - ( res == 0 ), 0 ) ; i++ ) printf ( "0" );
70 for ( int i = cnt ; i < num1 ; i++ ) printf ( "1" );
71 printf ( "\n" );
72 return 0;
73 }
K. King‘s Colors
给你一颗树有n个节点，用K种颜色染色，必要用完K种颜色，问方案数
第一次遇到容斥的题目，表示很不会
全部的方案为k*(k-1)^(n-1)-(k-1)*(k-1-1)^ (n-1)+ (k-2)*(k-2-1)^ (n-1)+……
我第一次看到这个式子很懵逼 后面解释是k*(k-1)^(n-1)这个包含了用了k-1的颜色的情况所以要减去
我第一次感觉减去这个就够了，但是这个式子还+上了k-2的颜色的情况 （我第一次想的时候k*(k-1)^(n-1)也包含了k-2的颜色的情况 为什么这里还要加上 直接减去不就好了）
然后我直接手写了一个4个节点和4种颜色的情况弄懂了
所有情况4*3*3*3 – 3*2*2*C(4,3)+C(4,2)*2*1*1*1
看3*2*2*C(4,3)这一项 选择3种颜色的方案数，这个里面包含了选择2种颜色的情况
假设抽出这3种颜色这里的两种颜色为
1 2 3 为 （1，2）（1，3），（2，3）
1 3 4 为 （1，3）（1，4），（3，4）
1 2 4 为 （1，2）（1，4），（2，4）
2 3 4 为 （2，3）（2，4），（3，4）
然后你会发现用了两种颜色的刚好多被减了一次，所以要加上
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define pi acos(-1.0)
13 #define eps 1e-9
14 #define fi first
15 #define se second
16 #define rtl rt<<1
17 #define rtr rt<<1|1
18 #define bug printf("******\n")
19 #define mem(a,b) memset(a,b,sizeof(a))
20 #define name2str(x) #x
21 #define fuck(x) cout<<#x" = "<<x<<endl
22 #define f(a) a*a
23 #define sf(n) scanf("%d", &n)
24 #define sff(a,b) scanf("%d %d", &a, &b)
25 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
26 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
27 #define pf printf
28 #define FRE(i,a,b) for(i = a; i <= b; i++)
29 #define FREE(i,a,b) for(i = a; i >= b; i--)
30 #define FRL(i,a,b) for(i = a; i < b; i++)+
31 #define FRLL(i,a,b) for(i = a; i > b; i--)
32 #define FIN freopen("data.txt","r",stdin)
33 #define gcd(a,b) __gcd(a,b)
34 #define lowbit(x) x&-x
35 #define rep(i,a,b) for(int i=a;i<b;++i)
36 #define per(i,a,b) for(int i=a-1;i>=b;--i)
37 using namespace std;
38 typedef long long LL;
39 typedef unsigned long long ULL;
40 const int INF = 0x3f3f3f3f;
41 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
42 const int maxn = 2505;
43 const int mod = 1e9 + 7;
44 int n, k, C[maxn][maxn];
45 LL exp_mod ( LL a, LL b ) {
46 LL ret = 1;
47 while ( b ) {
48 if ( b & 1 ) ret = ret * a % mod;
49 a = a * a % mod;
50 b = b >> 1;
51 }
52 return ret % mod;
53 }
54 int main() {
55 C[0][0] = 1, C[0][1] = 0;
56 for ( int i = 1 ; i <= 2500 ; i++ ) {
57 C[i][0] = 1, C[i][i + 1] = 0;
58 for ( int j = 1 ; j <= i ; j++ )
59 C[i][j] = ( C[i - 1][j - 1] + C[i - 1][j] ) % mod;
60 }
61 sff ( n, k );
62 for ( int i = 1, u ; i < n ; i++ ) sf ( u );
63 LL ans = 0;
64 for ( int i = k ; i >= 2 ; i-- )
65 if ( ( k - i ) % 2 == 0 ) ans = ( ans + 1LL * i * exp_mod ( i - 1, n - 1 ) % mod * C[k][i] % mod ) % mod;
66 else ans = ( ans - 1LL * i * exp_mod ( i - 1, n - 1 ) % mod * C[k][i] % mod + mod ) % mod;
67 printf ( "%lld\n", ans );
68 }
原文地址：https://www.cnblogs.com/qldabiaoge/p/10527219.html
时间： 2024-10-31 11:40:07