A题。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <cmath>
6 #include <vector>
7
8 using namespace std;
9
10 #define LL long long
11 #define eps 1e-8
12 #define inf 0x3f3f3f3f
13 #define lson l, m, rt << 1
14 #define rson m+1, r, rt << 1 | 1
15 #define mnx 500
16
17 struct node{
18 int v, id;
19 bool operator < ( const node &b ) const {
20 return v < b.v;
21 }
22 }a[mnx];
23 int b[mnx];
24 int main(){
25 int n, k;
26 scanf( "%d %d", &n, &k );
27 for( int i = 1; i <= n; ++i ){
28 scanf( "%d", &a[i].v );
29 a[i].id = i;
30 }
31 sort( a + 1, a + 1 + n );
32 int i = 1, ans = 0, sum = 0;
33 while( sum + a[i].v <= k && i <= n ){
34 b[ans++] = a[i].id, sum += a[i++].v;
35 }
36 printf( "%d\n", ans );
37 for( i = 0; i < ans; ++i )
38 printf( "%d ", b[i] );
39 return 0;
40 }
B题。。注意有可能 爆int
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <cmath>
6 #include <vector>
7
8 using namespace std;
9
10 #define LL long long
11 #define eps 1e-8
12 #define inf 0x3f3f3f3f
13 #define lson l, m, rt << 1
14 #define rson m+1, r, rt << 1 | 1
15 #define mnx 500
16
17 int main(){
18 double r, x, y, xx, yy;
19 scanf( "%lf%lf%lf%lf%lf", &r, &x, &y, &xx, &yy );
20 double dis = sqrt( ( xx - x ) * ( xx - x ) + ( yy - y ) * ( yy - y ) );
21 int ans = ceil( dis / 2 / r );
22 cout << ans << endl;
23 }
C题。。把它当做一棵满二叉树来看，根节点是1，然后第二层是2,3， 第三层是4,5,6,7。。先计算出第h层第n个节点在整个满二叉树是第cur个，然后一直向上得到父亲节点，一直到根节点。。计算的话，因为是LRLRLR这样走的，如果a[i+1] %2 == a[i]%2的话，就要加上某一边的子树的所有节点b[h+1-i]，不同的话就直接ans++就好。。具体自己画图加结合代码理解一下
1 include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <cmath>
5 #include <vector>
6
7 using namespace std;
8
9 #define LL long long
10 #define eps 1e-8
11 #define inf 0x3f3f3f3f
12 #define lson l, m, rt << 1
13 #define rson m+1, r, rt << 1 | 1
14 #define mnx 500
15
16 LL a[mnx], b[mnx];
17 int main(){
18 LL h, n;
19 cin >> h >> n;
20 LL L = 1;
21 for( int i = 0; i < h; ++i )
22 L <<= 1;
23 b[0] = 1;
24 for( int i = 1; i <= h; ++i )
25 b[i] = b[i-1] * 2;
26 LL cur = L + n - 1;
27 int m = h + 1;
28 while( cur ){
29 a[m--] = cur;
30 cur >>= 1;
31 }
32 LL ans = 0;
33 for( int i = 1; i < h+1; ++i ){
34 if( a[i+1] % 2 == a[i] % 2 )
35 ans += b[h+1-i];
36 else ans++;
37 }
38 cout << ans << endl;
39
D题。。不会，数位dp，复制别人的题解
给定n，k，m
求有多少个恰好为n位的整数，且这个整数的某个后缀能整除k，答案模m
思路：
因为是求后缀，所以从低位往高位构造。
dp[i][j][0] 表示长度为i位的整数，模k结果为j，且不存在某个后缀能整除k的个数
dp[i][j][1] 表示长度为i位的整数，模k结果为j，且存在某个后缀能整除k的个数
然后转移就在某个状态前加一个数字，判断一下加上后能否被k整除来决定加上数字后到达的状态
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <cmath>
6 #include <vector>
7 #include <map>
8 #include <queue>
9
10 using namespace std;
11
12 #define LL long long
13 #define eps 1e-8
14 #define inf 0x3f3f3f3f
15 #define lson l, m, rt << 1
16 #define rson m+1, r, rt << 1 | 1
17 #define mnx 1050
18
19 LL dp[mnx][120][2], len[mnx], mod;
20 int n, k;
21 void add( LL &x, LL y ){
22 x = ( x + y ) % mod;
23 }
24 int main(){
25 cin >> n >> k >> mod;
26 dp[0][0][0] = 1, len[0] = 1;
27 for( int i = 1; i < mnx; ++i )
28 len[i] = len[i-1] * 10 % k;
29 for( int i = 0; i < n; ++i ){
30 for( int j = 0; j < k; ++j ){
31 for( int v = 0; v < 10; ++v ){
32 if( i == n-1 && !v ) continue;
33 LL val = ( v * len[i] + j ) % k;
34 if( !val && v )
35 add( dp[i+1][0][1], dp[i][j][0] );
36 else
37 add( dp[i+1][val][0], dp[i][j][0] );
38 add( dp[i+1][val][1], dp[i][j][1] );
39 }
40 }
41 }
42 LL ans = 0;
43 for( int i = 0; i < k; ++i )
44 ans = ( ans + dp[n][i][1] ) % mod;
45 cout << ans << endl;
46 return 0;
47 }
Ｅ题。。最短路，感觉挺水的，把最短路上所有为0的边改为1，把不是最短路的边所有为1的变为0。。
表示自己的代码写的好挫。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <cmath>
6 #include <vector>
7 #include <map>
8 #include <queue>
9
10 using namespace std;
11
12 #define LL long long
13 #define eps 1e-8
14 #define inf 0x3f3f3f3f
15 #define lson l, m, rt << 1
16 #define rson m+1, r, rt << 1 | 1
17 #define mnx 200500
18
19 int fst[mnx], nxt[mnx], vv[mnx], work[mnx], e, dis[mnx], sum[mnx], n, fa[mnx];
20 bool vis[mnx];
21 map< pair<int,int>, int > mp;
22 void add( int u, int v, int c ){
23 vv[e] = v, nxt[e] = fst[u], work[e] = c, fst[u] = e++;
24 }
25 struct edge{
26 int v, u, d;
27 edge(){}
28 edge( int u, int v, int d ) : u(u), v(v), d(d) {}
29 bool operator < ( const edge &b ) const {
30 return d > b.d;
31 }
32 }ee[mnx];
33 void dij(){
34 memset( fa, -1, sizeof fa );
35 fa[1] = 1;
36 memset( dis, 0x3f, sizeof dis );
37 priority_queue<edge> q;
38 dis[1] = 0;
39 edge p;
40 p.u = 1, p.d = 0;
41 q.push( p );
42 while( !q.empty() ){
43 p = q.top(); q.pop();
44 int u = p.u;
45 if( vis[u] ) continue;
46 vis[u] = 1;
47 for( int i = fst[u]; i != -1; i = nxt[i] ){
48 int v = vv[i], w = work[i];
49 if( dis[v] > dis[u] + 1 || ( dis[v] == dis[u] + 1 && sum[v] < sum[u] + w ) ){
50 dis[v] = dis[u] + 1;
51 sum[v] = sum[u] + w;
52 edge pp;
53 pp.u = v, pp.d = dis[v];
54 q.push( pp );
55 fa[v] = u;
56 }
57 }
58 }
59 //cout << dis[n] << endl;
60 }
61 void dfs( int u ){
62 vis[u] = 1;
63 if( fa[u] != u ){
64 dfs( fa[u] );
65 }
66 }
67 int main(){
68 int m, all = 0;
69 scanf( "%d%d",&n, &m );
70 memset( fst, -1, sizeof fst );
71 for( int i = 0; i < m; ++i ){
72 int u, v, c;
73 scanf( "%d%d%d", &u, &v, &c );
74 add( u, v, c );
75 add( v, u, c );
76 ee[all++] = edge( u, v, c );
77 ee[all++] = edge( v, u, c );
78 }
79 dij();
80 memset( vis, 0, sizeof vis );
81 dfs( n );
82 int ans = 0;
83 for( int i = 1; i <= n; ++i ){
84 int u = i;
85 if( !vis[i] ){
86 for( int j = fst[u]; j != -1; j = nxt[j] ){
87 int v = vv[j];
88 if( mp.find( make_pair( u, v ) ) == mp.end() && mp.find( make_pair( v, u ) ) == mp.end() ){
89 if( work[j] ) ans++;
90 mp[make_pair(u,v)]++;
91 }
92 }
93 }
94 else{
95 for( int j = fst[u]; j != -1; j = nxt[j] ){
96 int v = vv[j];
97 if( v == fa[u] ){
98 if( !work[j] ) ans++;
99 mp[make_pair(u,v)]++;
100 }
101 }
102 }
103 }
104 for( int i = 1; i <= n; ++i ){
105 int u = i;
106 if( vis[i] ){
107 for( int j = fst[u]; j != -1; j = nxt[j] ){
108 int v = vv[j];
109 if( mp.find( make_pair(u,v) ) == mp.end() && mp.find( make_pair(v,u) ) == mp.end() ){
110 if( work[j] ) ans++;
111 mp[make_pair(u,v)]++;
112 }
113 }
114 }
115 }
116 cout << ans << endl;
117 mp.clear();
118 for( int i = 1; i <= n; ++i ){
119 int u = i;
120 if( !vis[i] ){
121 for( int j = fst[u]; j != -1; j = nxt[j] ){
122 int v = vv[j];
123 if( mp.find( make_pair( u, v ) ) == mp.end() && mp.find( make_pair( v, u ) ) == mp.end() ){
124 if( work[j] )
125 printf( "%d %d %d\n", min(u,v),max(u,v), 0 );
126 mp[make_pair(u,v)]++;
127 }
128 }
129 }
130 else{
131 for( int j = fst[u]; j != -1; j = nxt[j] ){
132 int v = vv[j];
133 if( v == fa[u] ){
134 if( !work[j] )
135 printf( "%d %d %d\n", min(u,v),max(u,v), 1 );
136 mp[make_pair(u,v)]++;
137 }
138 }
139 }
140 }
141 for( int i = 1; i <= n; ++i ){
142 int u = i;
143 if( vis[i] ){
144 for( int j = fst[u]; j != -1; j = nxt[j] ){
145 int v = vv[j];
146 if( mp.find( make_pair(u,v) ) == mp.end() && mp.find( make_pair(v,u) ) == mp.end() ){
147 if( work[j] )
148 printf( "%d %d %d\n", min(u,v),max(u,v), 0 );
149 mp[make_pair(u,v)]++;
150 }
151 }
152 }
153 }
154 return 0;
155 }
时间： 2024-10-05 15:57:26