标签：ace ++ cstring make 就是 print pair mes main
岛上有三个物种：剪刀$s$、石头$r$、布$p$
其中剪刀能干掉布，布能干掉石头，石头能干掉剪刀
每天会从这三个物种中发生一场战争(也就是说其中的一个会被干掉)
问最后仅有$s/r/p$物种生存的概率
还是想复杂了啊，我列的状态时$f[i][j], g[i][j],t[i][j]$分别表示第$i$天，$j$个$s, r, p$活着的概率
然而转移了一下午也没转移出来。。
标算比我简单的多，直接设$f[i][j][k]$表示剩下$i$个$s$，$j$个$r$，$k$个$p$的概率
然后记忆化搜索一下
/* */ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<map> #include<vector> #include<set> #include<queue> #include<cmath> //#include<ext/pb_ds/assoc_container.hpp> //#include<ext/pb_ds/hash_policy.hpp> #define Pair pair<int, int> #define MP(x, y) make_pair(x, y) #define fi first #define se second #define int long long #define LL long long #define ull unsigned long long #define rg register #define pt(x) printf("%d ", x); //#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++) //char buf[(1 << 22)], *p1 = buf, *p2 = buf; //char obuf[1<<24], *O = obuf; //void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + ‘0‘;} //#define OS *O++ = ‘ ‘; using namespace std; //using namespace __gnu_pbds; const int MAXN = 1e6 + 10, INF = 1e9 + 10, mod = 1e9 + 7; const double eps = 1e-9; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();} while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar(); return x * f; } double a, b, c; double f[101][101][101]; double calca(int a, int b, int c) { if(f[a][b][c]) return f[a][b][c]; if(a == 0) return f[a][b][c] = 0; if(a && (!b) && (!c)) return f[a][b][c] = 1; double down = a * b + b * c + a * c, ans = 0; if(a && b) ans += (double) a * b / down * calca(a, b - 1, c); if(b && c) ans += (double) b * c / down * calca(a, b, c - 1); if(a && c) ans += (double) a * c / down * calca(a - 1, b, c); return f[a][b][c] = ans; } double calcb(int a, int b, int c) { if(b == 0) return f[a][b][c] = 0; if((!a) && b && (!c)) return f[a][b][c] = 1; if(f[a][b][c]) return f[a][b][c]; double down = a * b + b * c + a * c, ans = 0; if(a && b) ans += (double) a * b / down * calcb(a, b - 1, c); if(b && c) ans += (double) b * c / down * calcb(a, b, c - 1); if(a && c) ans += (double) a * c / down * calcb(a - 1, b, c); return f[a][b][c] = ans; } main() { a = read(); b = read(); c = read(); double a1 = 0, a2 = 0; printf("%.10lf ", a1 = calca(a, b, c)); memset(f, 0, sizeof(f)); printf("%.10lf ", a2 = calcb(a, b, c)); printf("%.10lf", 1 - a1 - a2); return 0; } /* 2 2 1 1 1 2 1 1 */
标签：ace ++ cstring make 就是 print pair mes main
原文地址：https://www.cnblogs.com/zwfymqz/p/9574442.html
