HDU 1085 Holding Bin-Laden Captive! 母函数
2014-08-07 10:36
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这次变成了每一种硬币有数量限制的情况了,用母函数显然可以十分方便的解决
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack> using namespace std; typedef long long LL; const int maxn = 1000 + 2 * 1000 + 5 * 1000 + 5; int c1[maxn],c2[maxn],lim[3],val[3] = {1,2,5}; int main() { while(scanf("%d%d%d",&lim[0],&lim[1],&lim[2]),lim[0] + lim[1] + lim[2]) { int maxval = lim[0] + 2 * lim[1] + 5 * lim[2]; memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); for(int i = 0;i <= lim[0];i++) { c1[i] = 1; } for(int i = 1;i < 3;i++) { for(int j = 0;j <= lim[i];j++) { for(int k = 0;k + j * val[i] <= maxval;k++) { c2[k + j * val[i]] += c1[k]; } } for(int i = 0;i <= maxval;i++) { c1[i] = c2[i]; c2[i] = 0; } } int ans = 0; for(int i = 0;i <= maxval + 10;i++) if(!c1[i]) { ans = i; break; } printf("%d\n",ans); } return 0; }
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