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题意：有n个人，要分成k支队，问你最多能够进行多少场比赛。。贪心，尽量平均分，就能够进行最多场的比赛。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <vector> 6 #include <algorithm> 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 10100 16 #define mod 1000000007 17 18 int a[mnx]; 19 int main(){ 20 int n, m, cas; 21 scanf( "%d", &cas ); 22 while( cas-- ){ 23 scanf( "%d%d", &n, &m ); 24 int tmp = n / m; 25 for( int i = 1; i <= m; ++i ){ 26 a[i] = tmp; 27 } 28 tmp = n % m; 29 for( int i = 1; i <= tmp; ++i ){ 30 a[i]++; 31 } 32 for( int i = 1; i <= m; ++i ){ 33 a[i] = a[i-1] + a[i]; 34 } 35 LL ans = 0; 36 for( int i = 1; i <= m; ++i ){ 37 ans += (LL)( a[m] - a[i] ) * ( a[i] - a[i-1] ); 38 } 39 cout << ans << endl; 40 } 41 return 0; 42 }
题意：给你一个既不能被2又不能被5整除的数n，问你长度为多少的只含1的数能够整除 n。。
做法：数据比较小，暴力找就好了，肯定能够找到。前几天做过问 长度为多少的只含8的数能够整除n，那道题的数据比较大，用欧拉函数搞。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 #include <queue> 8 9 using namespace std; 10 11 #define LL long long 12 #define eps 1e-6 13 #define inf 0x3f3f3f3f 14 #define mod 1000 15 #define MP make_pair 16 #define mnx 10050 17 18 int main(){ 19 int n; 20 while( scanf( "%d", &n ) != EOF ){ 21 int tmp = 1, ans = 1; 22 while( tmp % n ){ 23 ans++; 24 tmp = ( tmp % n ) * 10 + 1; 25 } 26 printf( "%d\n", ans ); 27 } 28 return 0; 29 }
题意：给你一条弦还有圆弧的长度，问你 弦的中心到圆弧中心的距离。
做法：二分答案就好了。最开始做的时候二分角度没跑出来，以为不是单调的，后来爬山又跪了，心好塞。没想到是单调的，二分答案就好了。还有poj交G++的时候浮点数要用.3f
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 #include <queue> 8 9 using namespace std; 10 11 #define LL long long 12 #define eps 1e-8 13 #define inf 0x3f3f3f3f 14 #define mod 1000 15 #define MP make_pair 16 #define mnx 25 17 18 int main(){ 19 double seg, c, n, len; 20 while( scanf( "%lf%lf%lf", &seg, &n, &c ) != EOF && seg >= 0 ){ 21 len = seg + seg * c * n; 22 double l = 0, r = seg/2; 23 while( r - l > eps ){ 24 double m = ( r + l ) / 2; 25 double R = ( m*m + (seg/2)*(seg/2) ) / (2*m); 26 double a = asin( seg / 2 / R ); 27 double len2 = 2 * a * R; 28 if( len2 > len ) 29 r = m; 30 else l = m; 31 } 32 printf( "%.3lf\n", l ); 33 } 34 return 0; 35 }
看别人的题解吧，这里。不会做。现在好像有点明白了，树形dp + 背包。。C++能过，但G++不知道为什么TLE了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 #include <queue> 8 9 using namespace std; 10 11 #define LL long long 12 #define eps 1e-6 13 #define inf 0x3f3f3f3f 14 #define mod 1000 15 #define MP make_pair 16 #define mnx 20050 17 18 int n, s, k, dp[mnx][20]; 19 int vv[mnx], fst[mnx], cost[mnx], nxt[mnx], e; 20 void add( int u, int v, int c ){ 21 vv[e] = v, nxt[e] = fst[u], cost[e] = c, fst[u] = e++; 22 } 23 void dfs( int u, int fa ){ 24 for( int i = fst[u]; i != -1; i = nxt[i] ){ 25 int v = vv[i], c = cost[i]; 26 if( v == fa ) continue; 27 dfs( v, u ); 28 for( int j = k; j >= 0; --j ){ 29 dp[u][j] += dp[v][0] + 2 * c; 30 for( int jj = 1; jj <= j; ++jj ){ 31 dp[u][j] = min( dp[u][j], dp[u][j-jj] + dp[v][jj] + jj * c ); 32 } 33 } 34 } 35 } 36 int main(){ 37 while( scanf( "%d%d%d", &n, &s, &k ) != EOF ){ 38 memset( fst, -1, sizeof fst ); 39 memset( dp, 0, sizeof dp ); 40 e = 0; 41 int u, v, c; 42 for( int i = 0; i < n-1; ++i ){ 43 scanf( "%d%d%d", &u, &v, &c ); 44 add( u, v, c ); 45 add( v, u, c ); 46 } 47 dfs( s, -1 ); 48 printf( "%d\n", dp[s][k] ); 49 } 50 return 0; 51 }
题意：给你一个图，让你把图联通，同时使权值最大的边尽量小。
做法：最小生成树的kruskal算法，题目有spj的。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <vector> 6 #include <algorithm> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-8 12 #define lson l, m, rt<<1 13 #define rson m+1, r, rt<<1|1 14 #define mnx 100100 15 16 struct edge{ 17 int u, v, c; 18 bool operator < ( const edge &b ) const { 19 return c < b.c; 20 } 21 void input(){ 22 scanf( "%d%d%d", &u, &v, &c ); 23 } 24 }e[mnx]; 25 int fa[mnx]; 26 int find( int x ){ 27 if( fa[x] != x ) 28 fa[x] = find( fa[x] ); 29 return fa[x]; 30 } 31 int uu[mnx], vv[mnx]; 32 int main(){ 33 int n, m; 34 while( scanf( "%d%d", &n, &m ) != EOF ){ 35 for( int i = 0; i <= n; ++i ) 36 fa[i] = i; 37 for( int i = 0; i < m; ++i ){ 38 e[i].input(); 39 } 40 sort( e, e + m ); 41 int ans = 0, cnt = 0; 42 for( int i = 0; i < m; ++i ){ 43 int u = e[i].u, v = e[i].v, c = e[i].c; 44 if( find(u) != find(v) ){ 45 fa[find(v)] = find(u); 46 ans = c; 47 uu[cnt] = u, vv[cnt++] = v; 48 } 49 } 50 printf( "%d\n%d\n", ans, cnt ); 51 for( int i = 0; i < cnt; ++i ) 52 printf( "%d %d\n", uu[i], vv[i] ); 53 } 54 return 0; 55 }
题意：自己看吧。
做法：模拟题，细心一点就可以过了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 #include <queue> 8 9 using namespace std; 10 11 #define LL long long 12 #define eps 1e-6 13 #define inf 0x3f3f3f3f 14 #define mod 1000 15 #define MP make_pair 16 #define mnx 105 17 18 int n, m, tot, cnt, a[mnx*1000], cur, L[30], step; 19 char ch[mnx][mnx]; 20 int dx[5] = { -1, 0, 1, 0 }, dy[5] = { 0, 1, 0, -1 }, D; 21 bool out( int x, int y ){ 22 if( x < 0 || y < 0 || x >= n || y >= m ) return 1; 23 return 0; 24 } 25 int gao( int x, int y ){ 26 if( out( x, y ) ) return 1; 27 char c = ch[x][y]; 28 if( c == ‘?‘ ){ 29 cur = a[cnt++]; 30 if( cnt >= tot ) cnt = tot - 1; 31 } 32 else if( c >= ‘A‘ && c <= ‘Z‘ ) 33 swap( cur, L[c-‘A‘] ); 34 else if( c == ‘!‘ ){ 35 printf( "%d\n", cur ); 36 cur = 0; 37 } 38 else if( c == ‘^‘ ) 39 D = 0; 40 else if( c == ‘>‘ ) 41 D = 1; 42 else if( c == ‘v‘ ) 43 D = 2; 44 else if( c == ‘<‘ ) 45 D = 3; 46 else if( c == ‘+‘ ) 47 cur++; 48 else if( c == ‘-‘ ) 49 cur--; 50 else if( c == ‘@‘ ){ 51 if( cur == 0 ) 52 D = ( D + 4 - 1 ) % 4; 53 else D = ( D + 1 ) % 4; 54 } 55 else if( c == ‘#‘ ) 56 return -1; 57 return 0; 58 } 59 int main(){ 60 scanf( "%d%d", &n, &m ); 61 for( int i = 0; i < n; ++i ) 62 scanf( "%s", ch[i] ); 63 scanf( "%d", &tot); 64 for( int i = 0; i < tot; ++i ) 65 scanf( "%d", &a[i] ); 66 int x = 0, y = 0; 67 D = 1, step = 0; 68 while( 1 ){ 69 int ok = gao( x, y ); 70 if( ok == 1 ){ 71 puts( "RUNTIME ERROR" ); break; 72 } 73 if( ok == -1 ){ 74 break; 75 } 76 if( abs(cur) > 100000 ){ 77 puts( "OVERFLOW ERROR" ); break; 78 } 79 step++; 80 if( step >= 1000000 ){ 81 puts( "TIME LIMIT EXCEEDED" ); break; 82 } 83 x = x + dx[D], y = y + dy[D]; 84 } 85 return 0; 86 }
题意：有一个robot要清除地图上所有不干净的地方，问你最短路要多少。
做法：bfs + TSP问题。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 #include <queue> 8 9 using namespace std; 10 11 #define LL long long 12 #define eps 1e-8 13 #define inf 0x3f3f3f3f 14 #define mod 1000 15 #define MP make_pair 16 #define mnx 25 17 18 char ch[mnx][mnx]; 19 int dx[5] = { 0, 1, -1, 0 }, dy[5] = { 1, 0, 0, -1 }; 20 bool vis[mnx][mnx]; 21 struct node{ 22 int u, v, dis; 23 node( int u = 0, int v = 0, int dis = 0 ) : u(u), v(v), dis(dis) {} 24 }; 25 queue<node> q; 26 int x[mnx], y[mnx], g[mnx][mnx], dis[mnx][mnx], cnt, n, m; 27 bool in( int ax, int ay ){ 28 if( ax < 0 || ay < 0 || ax >= n || ay >= m || ch[ax][ay] == ‘x‘ ) 29 return 0; 30 return 1; 31 } 32 void bfs( int k, int cx, int cy ){ 33 memset( vis, 0, sizeof(vis) ); 34 vis[cx][cy] = 1; 35 dis[k][k] = 0; 36 q.push( node(cx, cy, 0) ); 37 while( !q.empty() ){ 38 node vv = q.front(); 39 int gx = vv.u, gy = vv.v, d = vv.dis; 40 q.pop(); 41 for( int i = 0; i < 4; ++i ){ 42 int ax = gx + dx[i], ay = gy + dy[i]; 43 if( in( ax, ay ) && !vis[ax][ay] ){ 44 if( g[ax][ay] != -1 ) 45 dis[k][g[ax][ay]] = d + 1; 46 vis[ax][ay] = 1; 47 q.push( node(ax, ay, d+1) ); 48 } 49 } 50 } 51 } 52 int dp[15000][15]; 53 void solve(){ 54 for( int i = 0; i < cnt; ++i ) 55 bfs( i, x[i], y[i] ); 56 cnt--; 57 for( int v = 0; v < (1<<cnt); ++v ){ 58 for( int i = 1; i <= cnt; ++i ){ 59 if( v & (1<<(i-1)) ){ 60 if( v == (1<<(i-1)) ) 61 dp[v][i] = dis[0][i]; 62 else{ 63 dp[v][i] = inf; 64 for( int j = 1; j <= cnt; ++j ){ 65 if( v & (1<<(j-1)) && j != i ) 66 dp[v][i] = min( dp[v][i], dp[v^(1<<(i-1))][j] + dis[j][i] ); 67 } 68 } 69 } 70 } 71 } 72 int ans = inf; 73 for( int i = 1; i <= cnt; ++i ){ 74 ans = min( dp[((1<<cnt)-1)][i], ans ); 75 } 76 if( ans == inf ) puts( "-1" ); 77 else printf( "%d\n", ans ); 78 } 79 int main(){ 80 //freopen( "tt.txt", "r", stdin ); 81 while( scanf( "%d%d", &m, &n ) != EOF && ( m && n ) ){ 82 memset( g, -1, sizeof(g) ); 83 memset( dis, 0x3f, sizeof(dis) ); 84 cnt = 1; 85 for( int i = 0; i < n; ++i ){ 86 scanf( "%s", ch[i] ); 87 for( int j = 0; j < m; ++j ){ 88 if( ch[i][j] == ‘*‘ ){ 89 x[cnt] = i, y[cnt] = j, g[i][j] = cnt++; 90 } 91 if( ch[i][j] == ‘o‘ ) 92 x[0] = i, y[0] = j, g[i][j] = 0; 93 } 94 } 95 solve(); 96 } 97 return 0; 98 }
题意：有w个mice和b个mice，公主 和 龙 轮流摸，谁先摸到白的谁就赢，问你公主获胜的概率。龙每次摸完之后，如果没有赢的话，就会有一只mice会逃走（白色的或者黑色的）
做法：dp[i][j]表示有i个白mice和j个黑mice，公主赢的概率。根据题意，公主赢的概率有3种情况。
dp[i][j] = i*1.0 / sum; （当前公主赢的概率）
dp[i][j] += ( dp[i-1][j-2] * (j/sum * (j-1.0)/(sum-1.0) * (i/(sum-2.0)) ) ); （龙没有赢，逃走的是白mice）
dp[i][j] += ( dp[i][j-3] * (j/sum * (j-1.0)/(sum-1.0) * (j-2.0)/(sum-2.0) ) ); （龙没有赢，逃走的是黑mice）
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <vector> 6 #include <algorithm> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-12 12 #define lson l, m, rt<<1 13 #define rson m+1, r, rt<<1|1 14 #define mnx 1100 15 #define Pi acos( -1.0 ) 16 17 double dp[mnx][mnx]; 18 int main(){ 19 for( int i = 1; i < mnx; ++i ){ 20 dp[i][0] = 1.0; 21 } 22 dp[1][1]; 23 int w, b; 24 while( scanf( "%d%d", &w, &b ) != EOF ){ 25 for( int i = 1; i <= w; ++i ){ 26 for( int j = 1; j <= b; ++j ){ 27 double sum = i + 0.0 + j; 28 dp[i][j] = i*1.0 / sum; 29 if( j >= 2 ){ 30 dp[i][j] += ( dp[i-1][j-2] * (j/sum * (j-1.0)/(sum-1.0) * (i/(sum-2.0)) ) ); 31 } 32 if( j >= 3 ){ 33 dp[i][j] += ( dp[i][j-3] * (j/sum * (j-1.0)/(sum-1.0) * (j-2.0)/(sum-2.0) ) ); 34 } 35 } 36 } 37 printf( "%.10lf\n", dp[w][b] ); 38 } 39 return 0; 40 }
题意：给你n个单词，问你这n个单词里面有哪些个是可以由两个单词组成的。
做法：trie树 or 哈希。还以为单词很长，一个个分割会超时，没想到单词并不长。
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 mod 1000 14 #define mnx 200100 15 16 char ch[mnx][50]; 17 int son[mnx][26], cnt, val[mnx], all; 18 void insert( char *s ){ 19 int m = strlen( s ), t = 0; 20 for( int i = 0; i < m; ++i ){ 21 int c = s[i] - ‘a‘; 22 if( son[t][c] == 0 ){ 23 son[t][c] = ++cnt; 24 memset( son[cnt], 0, sizeof(son[cnt]) ); 25 val[cnt] = 0; 26 } 27 t = son[t][c]; 28 } 29 val[t] = 1; 30 } 31 bool find( char *s ){ 32 int m = strlen( s ), t = 0; 33 for( int i = 0; i < m; ++i ){ 34 int c = s[i] - ‘a‘; 35 if( son[t][c] == 0 ) 36 return false; 37 t = son[t][c]; 38 } 39 return val[t]; 40 } 41 char s[mnx]; 42 bool gao( int c, int u, int v ){ 43 int tot = 0; 44 for( int i = u; i <= v; ++i ){ 45 s[tot++] = ch[c][i]; 46 } 47 s[tot] = ‘\0‘; 48 //cout << s << endl; 49 if( find( s ) ) return 1; 50 return 0; 51 } 52 int main(){ 53 cnt = all = 0; 54 while( scanf( "%s", ch[all] ) != EOF ){ 55 insert( ch[all] ); 56 all++; 57 } 58 for( int i = 0; i < all; ++i ){ 59 int n = strlen( ch[i] ); 60 for( int j = 0; j < n-1; ++j ){ 61 if( gao( i, 0, j ) && gao( i, j + 1, n-1 ) ){ 62 printf( "%s\n", ch[i] ); break; 63 } 64 } 65 } 66 return 0; 67 }
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原文地址：http://www.cnblogs.com/LJ-blog/p/4379694.html
