A题，n*m根木棍，相交放置，轮流取走相交的两根，最后谁不能行动，则输掉。
min(n,m)&1 为1则先取者赢。
B题，给定一个长度为n，且各不相同的数组，问能否通过交换连续一段L....R使得变成单调递增。
如果一开始就是递增的，那么直接输出L。。。R就是1 1，交换一个就行了；否则判断中间是否有且一段单调递减，且两端交换后使得数组递增。
代码：
1 //Template updates date: 20140718
2 #include <iostream>
3 #include <sstream>
4 #include <cstdio>
5 #include <climits>
6 #include <ctime>
7 #include <cctype>
8 #include <cstring>
9 #include <cstdlib>
10 #include <string>
11 #include <stack>
12 #include <set>
13 #include <map>
14 #include <cmath>
15 #include <vector>
16 #include <queue>
17 #include <algorithm>
18 #define esp 1e-6
19 #define inf 0x3f3f3f3f
20 #define pi acos(-1.0)
21 #define pb push_back
22 #define lson l, m, rt<<1
23 #define rson m+1, r, rt<<1|1
24 #define lowbit(x) (x&(-x))
25 #define mp(a, b) make_pair((a), (b))
26 #define bit(k) (1<<(k))
27 #define iin freopen("pow.in", "r", stdin);
28 #define oout freopen("pow.out", "w", stdout);
29 #define in freopen("solve_in.txt", "r", stdin);
30 #define out freopen("solve_out.txt", "w", stdout);
31 #define bug puts("********))))))");
32 #define Inout iin oout
33 #define inout in out
34
35 #define SET(a, v) memset(a, (v), sizeof(a))
36 #define SORT(a) sort((a).begin(), (a).end())
37 #define REV(a) reverse((a).begin(), (a).end())
38 #define READ(a, n) {REP(i, n) cin>>(a)[i];}
39 #define REP(i, n) for(int i = 0; i < (n); i++)
40 #define VREP(i, n, base) for(int i = (n); i >= (base); i--)
41 #define Rep(i, base, n) for(int i = (base); i < (n); i++)
42 #define REPS(s, i) for(int i = 0; (s)[i]; i++)
43 #define pf(x) ((x)*(x))
44 #define mod(n) ((n))
45 #define Log(a, b) (log((double)b)/log((double)a))
46 #define Srand() srand((int)time(0))
47 #define random(number) (rand()%number)
48 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)
49
50 using namespace std;
51 typedef long long LL;
52 typedef unsigned long long ULL;
53 typedef vector<int> VI;
54 typedef pair<int,int> PII;
55 typedef vector<PII> VII;
56 typedef vector<PII, int> VIII;
57 typedef VI:: iterator IT;
58 typedef map<string, int> Mps;
59 typedef map<int, int> Mpi;
60 typedef map<int, PII> Mpii;
61 typedef map<PII, int> Mpiii;
62
63 const int maxn = 100000 + 100;
64 int a[maxn];
65
66 int main() {
67
68 int n;
69 Rep(i, scanf("%d", &n), n+1) scanf("%d", a+i);
70 int l = 0, r = 0;
71 int ok = 0;
72 a[n+1] = inf;
73 Rep(i, 1, n+2) {
74 if(a[i] > a[i-1]) {
75 if(!l)
76 continue;
77 else {
78 if(ok && !r)
79 r = i-1;
80 }
81 } else if(a[i] < a[i-1]) {
82 if(!ok)
83 ok = 1, l = i-1;
84 if(r) {
85 cout<<"no"<<endl;
86 return 0;
87 }
88 } else {
89 cout<<"no"<<endl;
90 return 0;
91 }
92 }
93 if(l) {
94 if(a[l] < a[r+1] && a[r] > a[l-1]) {
95 cout<<"yes"<<endl;
96 cout<<l <<‘ ‘<<r<<endl;
97 } else {
98 cout<<"no"<<endl;
99 }
100 } else {
101 cout<<"yes"<<endl;
102 cout<<1<<‘ ‘<<1<<endl;
103 }
104 return 0;
105 }
写得挫。
C题，3个球队，总共n场比赛，已进行k场，知道两个球队之间赢得场数的绝对值，如|x1-x2| = d1, |x2-x3| = d2, 问剩下的比赛中是否存在一种可能使得三个队伍赢得场数相等。
分析：题意知道每场比赛只用知道谁赢就行 ，不用顾忌各种规则，前k场中分别可能情况：
令x = x2;
x+d1, x, x+d2;
x+d1, x, x-d2;
x-d1, x, x+d2;
x-d1, x, x-d1;
对应4种情况求出x，在求出x1,x2,x3,判断满足0<=xi<=n/3 && xi <= k;
就可以了。
代码：
1 //Template updates date: 20140718
2 #include <iostream>
3 #include <sstream>
4 #include <cstdio>
5 #include <climits>
6 #include <ctime>
7 #include <cctype>
8 #include <cstring>
9 #include <cstdlib>
10 #include <string>
11 #include <stack>
12 #include <set>
13 #include <map>
14 #include <cmath>
15 #include <vector>
16 #include <queue>
17 #include <algorithm>
18 #define esp 1e-6
19 #define inf 0x3f3f3f3f
20 #define pi acos(-1.0)
21 #define pb push_back
22 #define lson l, m, rt<<1
23 #define rson m+1, r, rt<<1|1
24 #define lowbit(x) (x&(-x))
25 #define mp(a, b) make_pair((a), (b))
26 #define bit(k) (1<<(k))
27 #define iin freopen("pow.in", "r", stdin);
28 #define oout freopen("pow.out", "w", stdout);
29 #define in freopen("solve_in.txt", "r", stdin);
30 #define out freopen("solve_out.txt", "w", stdout);
31 #define bug puts("********))))))");
32 #define Inout iin oout
33 #define inout in out
34
35 #define SET(a, v) memset(a, (v), sizeof(a))
36 #define SORT(a) sort((a).begin(), (a).end())
37 #define REV(a) reverse((a).begin(), (a).end())
38 #define READ(a, n) {REP(i, n) cin>>(a)[i];}
39 #define REP(i, n) for(int i = 0; i < (n); i++)
40 #define VREP(i, n, base) for(int i = (n); i >= (base); i--)
41 #define Rep(i, base, n) for(int i = (base); i < (n); i++)
42 #define REPS(s, i) for(int i = 0; (s)[i]; i++)
43 #define pf(x) ((x)*(x))
44 #define mod(n) ((n))
45 #define Log(a, b) (log((double)b)/log((double)a))
46 #define Srand() srand((int)time(0))
47 #define random(number) (rand()%number)
48 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)
49
50 using namespace std;
51 typedef long long LL;
52 typedef unsigned long long ULL;
53 typedef vector<int> VI;
54 typedef pair<int,int> PII;
55 typedef vector<PII> VII;
56 typedef vector<PII, int> VIII;
57 typedef VI:: iterator IT;
58 typedef map<string, int> Mps;
59 typedef map<int, int> Mpi;
60 typedef map<int, PII> Mpii;
61 typedef map<PII, int> Mpiii;
62
63 LL n, k, d1, d2;
64 int a[][2] = {{-1, 1} ,{-1, -1}, {1, 1}, {1, -1}};
65
66 int main() {
67 int T;
68 for(int t = scanf("%d", &T); t <= T; t++) {
69 scanf("%I64d%I64d%I64d%I64d", &n, &k, &d1, &d2);
70 if(n%3) {
71 puts("no");
72 continue;
73 }
74 int ok = 0;
75 LL nn;
76 REP(i, 4) {
77 nn = k;
78 LL tmp = (LL)a[i][0]*d1+(LL)a[i][1]*d2;
79 nn -= tmp;
80 if(nn < 0 || nn%3)
81 continue;
82 nn /= 3;
83 if(nn > n/3 || nn > k)
84 continue;
85 LL x1 = nn+(LL)a[i][0]*d1;
86 LL x2 = nn+(LL)a[i][1]*d2;
87 if(x1 < 0 || x1 > n/3 || x2 < 0 || x2 > n/3 || x1 > k || x2 > k)
88 continue;
89 ok = 1;
90 break;
91 }
92 puts(ok ? "yes" : "no");
93 }
94 return 0;
95 }
D题，给定仅有‘a‘，‘a‘组成的字符串，定义一种子串：通过合并连续的字母最后可以得到一个回文串。如”aaabbaaa"最后得到"aba"。
很容易想到一个字串首尾字符相等时就为这样的要求的特殊字串。
题目要求奇数长度和偶数长度这样的字串个数，统计每个字符分别用dp[0][s[i]-‘a‘], dp[1][s[i]-‘a‘],表示位于i及之前的奇数位置和偶数位置的s[i]字符个数，
这样偶数长度ans0 += dp[!(i&1)][s[i]-‘a‘], 奇数长度ans1 += dp[i&1][s[i]-‘a‘];
代码：
1 //Template updates date: 20140718
2 #include <iostream>
3 #include <sstream>
4 #include <cstdio>
5 #include <climits>
6 #include <ctime>
7 #include <cctype>
8 #include <cstring>
9 #include <cstdlib>
10 #include <string>
11 #include <stack>
12 #include <set>
13 #include <map>
14 #include <cmath>
15 #include <vector>
16 #include <queue>
17 #include <algorithm>
18 #define esp 1e-6
19 #define inf 0x3f3f3f3f
20 #define pi acos(-1.0)
21 #define pb push_back
22 #define lson l, m, rt<<1
23 #define rson m+1, r, rt<<1|1
24 #define lowbit(x) (x&(-x))
25 #define mp(a, b) make_pair((a), (b))
26 #define bit(k) (1<<(k))
27 #define iin freopen("pow.in", "r", stdin);
28 #define oout freopen("pow.out", "w", stdout);
29 #define in freopen("solve_in.txt", "r", stdin);
30 #define out freopen("solve_out.txt", "w", stdout);
31 #define bug puts("********))))))");
32 #define Inout iin oout
33 #define inout in out
34
35 #define SET(a, v) memset(a, (v), sizeof(a))
36 #define SORT(a) sort((a).begin(), (a).end())
37 #define REV(a) reverse((a).begin(), (a).end())
38 #define READ(a, n) {REP(i, n) cin>>(a)[i];}
39 #define REP(i, n) for(int i = 0; i < (n); i++)
40 #define VREP(i, n, base) for(int i = (n); i >= (base); i--)
41 #define Rep(i, base, n) for(int i = (base); i < (n); i++)
42 #define REPS(s, i) for(int i = 0; (s)[i]; i++)
43 #define pf(x) ((x)*(x))
44 #define mod(n) ((n))
45 #define Log(a, b) (log((double)b)/log((double)a))
46 #define Srand() srand((int)time(0))
47 #define random(number) (rand()%number)
48 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)
49
50 using namespace std;
51 typedef long long LL;
52 typedef unsigned long long ULL;
53 typedef vector<int> VI;
54 typedef pair<int,int> PII;
55 typedef vector<PII> VII;
56 typedef vector<PII, int> VIII;
57 typedef VI:: iterator IT;
58 typedef map<string, int> Mps;
59 typedef map<int, int> Mpi;
60 typedef map<int, PII> Mpii;
61 typedef map<PII, int> Mpiii;
62
63
64 const int maxn = 100000 + 100;
65 char s[maxn];
66 LL dp[2][2];
67 int main() {
68
69 scanf("%s", s);
70 LL ans1 = 0, ans2 = 0;
71 REPS(s, i) {
72 int x = s[i]-‘a‘;
73 int y = i&1;
74 dp[y][x]++;
75 ans1+=dp[y][x];
76 ans2+=dp[y^1][x];
77 }
78 cout<<ans2<<‘ ‘<<ans1<<endl;
79 return 0;
80 }
E题，n个箱子装有fi朵花，箱子内花完全一样，箱子之间花看做不同，求从中选出s朵花，有多少选法？
我也不会数论，不过还是硬着头皮看懂了。以后要慢慢积累啊
分析： 设第i个箱子选出xi朵花。得到x1+x2+x3.....xn = s, 且0<=xi<=fi.
那么选花方案数就是解的组数，即（1+x+x^2+x^3+.....x^f1)*................(1+x+x^2+x^3+x^4.........x^fn)
= (1-x^(f1+1))******(1-x^(fn+1)) *(1-x)^(-n)（等比数列）中x^s系数
然后就是求里面x^s的系数了。
这个先算出前面部分各次幂的系数诚意后面(1-x)^(-n)相应系数，使得总系数为s。
其中（1-x)^(-n)幂为k的系数为C(n+k-1, k) = C(n+k-1, n-1)由于n很小（n<=20).
所以可以通过枚举法求前面的系数，然后C(n+k-1,n-1)利用逆元公式。
代码：
Codeforces 258 Div2
时间： 2024-10-09 03:11:14