hdu 1398 Square Coins（母函数|完全背包）

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http://acm.hdu.edu.cn/showproblem.php?pid=1398

题意：有价值为1^2,2^2....7^2的硬币共17种，每种硬币都有无限个。问用这些硬币能够组成价值为n的钱数共有几种方案数。

母函数：

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>#define LL long long#define _LL __int64#define eps 1e-12#define PI acos(-1.0)using namespace std;int c1[310],c2[310];int main(){    int n;    for(int i = 0; i <= 300; i++)    {        c1[i] = 1;        c2[i] = 0;    }    for(int i = 2; i <= 17; i++)    {        for(int j = 0; j <= 300; j++)        {            for(int k = 0; k+j <= 300; k += i*i) //增量变为i*i                c2[k+j] += c1[j];        }        for(int j = 0; j <= 300; j++)        {            c1[j] = c2[j];            c2[j] = 0;        }    }    while(~scanf("%d",&n)&&n)    {        cout << c1[n] << endl;    }    return 0;}

完全背包：每种物品都有一定的价值i*i，数目有无限件，那么价值为v的背包能放下物品组合的种类为

f[v] = sum{f[v-k*val[i]] | c[i]*k <= V}。初始化f[0] = 1.

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>#define LL long long#define _LL __int64#define eps 1e-12#define PI acos(-1.0)using namespace std;int dp[300];int main(){int n;while(~scanf("%d",&n)&&n){memset(dp,0,sizeof(dp));dp[0] = 1;for(int i = 1; i <= 17; i++){for(int j = i*i; j <= n; j++)dp[j] += dp[j-i*i];}printf("%d\n",dp[n]);}return 0;}

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