标签:style blog color io os ar for div sp
题意 把数字T分成N个数的和,保证这N个数中最小的数大于P。求方案数目
另f[i][j]表示把i分成j个数的和的方案数
f[i][j]=f[i][j-1]+f[i-1][j-1]+f[i-2][j-1]+...f[0][j-1];
f[i-1][j]=f[i-1][j-1]+f[i-2][j-1]+...f[0][j-1];
两式做差 推出f[i][j]=f[i-1][j]+f[i][j-1];
那么ans=f[T-NP][N];
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} #define MAXN 500 LL f[MAXN][MAXN]; int N,T,P; void init() { for (int i=0;i<MAXN;i++) for (int j=0;j<MAXN;j++) f[i][j]=0; for (int i=1;i<=200;i++) f[0][i]=1; for (int i=1;i<=200;i++) for (int j=1;j<=200;j++) f[i][j]=f[i-1][j]+f[i][j-1]; } int main() { init(); int kase; scanf("%d",&kase); while (kase--) { scanf("%d%d%d",&N,&T,&P); if (T-N*P<0) {puts("0");continue;} else printf("%lld\n",f[T-N*P][N]); } return 0; }
标签:style blog color io os ar for div sp
原文地址:http://www.cnblogs.com/Commence/p/3983442.html