POJ 1942 Paths on a Grid
2010-12-31 10:22
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解题思路:组合数学
1->递推:C(m,n)=c(m-1,n)+C(m,n-1),时间复杂度O(m*n)
2->组合 代码
1->递推:C(m,n)=c(m-1,n)+C(m,n-1),时间复杂度O(m*n)
2->组合 代码
#include <iostream> using namespace std; int main() { __int64 i,m,n,ans; while(scanf("%I64u %I64u", &m, &n)&&m+n) { if(m>n)swap(m,n); for(ans=1,i=0;i<m;i++) ans=ans*(m+n-i)/(i+1); printf("%I64u\n",ans); } return 0; }
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