POJ-1942-Paths on a Grid-组合水题

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http://poj.org/problem?id=1942

求C(n+m,n)

这题数据有点特殊....

C（a,b）的b 取min(b,a-b)就能ac了。。。

求组合只能实打实边乘边除了。。用ll或double足矣

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <map>#include <set>#include <stack>#include <iostream>using namespace std;  __int64 inf=15;double eps=0.000001;    __int64 C(__int64 a,__int64 b){__int64 ans=1;if (b<a-b) b=a-b;__int64 i;  __int64 tmp=a-b;for (i=a;i>b;i--){ans*=i;while(tmp>1&& ans%tmp==0 ){ans/=tmp;tmp--;}}return ans;}int main() { __int64 n,m;while(scanf("%I64d%I64d",&n,&m)!=EOF){if (!n&&!m) break;__int64 ans=1;__int64 tmp=n+m; ans=C(tmp,n);  printf("%I64d\n",ans);}return 0;}

double：

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <map>#include <set>#include <stack>#include <iostream>using namespace std;  __int64 inf=15;double eps=0.000001;    __int64 C(__int64 a,__int64 b){double  ans=1;if (b<a-b) b=a-b;__int64 i;  __int64 tmp=a-b;for (i=a;i>b;i--){ans*=i;while(tmp>1&& (__int64)(ans)%tmp==0 ){ ans=ans/ (double)tmp;ans=(__int64)(ans+0.5);tmp--;}}return (__int64)(ans+0.5);}int main() { __int64 n,m;while(scanf("%I64d%I64d",&n,&m)!=EOF){if (!n&&!m) break;__int64 ans=1;__int64 tmp=n+m; ans=C(tmp,n);  printf("%I64d\n",ans);}return 0;}

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