[HDU 1085] Holding Bin-Laden Captive! 母函数或多重背包

来源：互联网发布：西安建筑软件培训班编辑：程序博客网时间：2024/06/05 03:14

http://acm.hdu.edu.cn/showproblem.php?pid=1085

题意：有三个硬币面值分别是1，2，5。输入三个硬币的个数，要求出这三个硬币不能组成的最小的面值。

思路：可以用母函数或者多重背包。

母函数代码（171ms）：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int coin[3] = {1, 2, 5};int nex[8005],temp[8005];int main(){    int num[3];    while(cin>>num[0]>>num[1]>>num[2] && (num[0] || num[1] || num[2]))    {        int maxn = num[0] * 1 + num[1] * 2 + num[2] * 5; //最高次幂        memset(nex, 0, sizeof(nex));        memset(temp, 0, sizeof(temp));        for(int i = 0; i <= num[0]; i++){            nex[i] = 1;        }        //母函数为(1+x+x^2+...x^num[0])(1+x^2+x^4+....x^2*num[1])(1+x^5+x^10+...+x^5*num[2])         for(int i = 1; i < 3; i++) //一共三项        {            for(int j = 0; j <= maxn; j++)  //x^j            {                for(int k = 0; k + j <= maxn && k <= coin[i] * num[i]; k += coin[i]){    //x^k                    temp[k+j] += nex[j];  //相乘后x^(k+j)的系数加上x^k的系数乘以x^j的系数                }            }           //更新            for(int j = 0; j <= maxn; j++){                nex[j] = temp[j];                temp[j] = 0;            }        }        for(int i = 0; i <= maxn+1; i++){            if(nex[i] == 0)            {                cout<<i<<endl;                break;            }        }    }    return 0;}

多重背包（0ms）：

#include<cmath>#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int dp[8005];void ZeroOnePack(int volume, int val, int vol) //01背包{    for(int i = volume; i >= vol; i--){        dp[i] = dp[i-vol] + val;    }}void CompletelyPack(int volume, int val, int vol) //完全背包{    for(int i = vol; i <= volume; i++){        dp[i] = dp[i-vol] + val;    }}void MultiplePack(int volume, int val, int vol, int counts)//多重背包{    if(vol * counts > volume){  //装不下当前物体，看作完全背包        CompletelyPack(volume, val, vol);    } else{        int mid = 1;        while(mid <= counts){  //将物体拆成1，2，4，.....  然后当成01背包            ZeroOnePack(volume, val*mid, vol*mid);            counts = counts - mid;            mid = mid << 1;        }        if(counts){            ZeroOnePack(volume, val*counts, vol*counts);  //剩下的        }    }}int main(){    int x, y, z;    while(cin>>x>>y>>z && (x || y || z))    {        memset(dp, 0, sizeof(dp));        int volume = x + 2 * y + 5 * z;        MultiplePack(volume, 1, 1, x);        MultiplePack(volume, 2, 2, y);        MultiplePack(volume, 5, 5, z);        for(int i = 1; i <= volume+1; i++){            if(dp[i] != i){                cout<<i<<endl;                break;            }        }    }    return 0;}

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