HDU 1059 Dividing(多重背包)

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多重背包判断 sum 为偶数的时候 sum/2 可达不可达。

还是很简单的。只要sum/2 那么 sum 一定可以分成两份。

多重背包解决。

#include <cstdio>#include <algorithm>#include <iostream>#include <cstring>#include <cmath>#include <cstdlib>#include <string>#include <map>#include <vector>#include <set>#include <queue>#include <stack>#include <cctype>using namespace std;typedef long long LL;typedef unsigned long long ULL;#define MAXN 10+10#define INF (1<<30)#define mod 123456789int main (){    int a[MAXN] = {0};    int kase = 0;    while(true){        int sum = 0;        for(int i = 1; i <= 6; i++){            scanf("%d",&a[i]);            sum += a[i]*i;        }        if(!sum)            break;        printf("Collection #%d:\n",++kase);        if(sum & 1){            printf("Can't be divided.\n\n");            continue;        }        int d[60000+10] = {0};        sum /= 2;        for(int i = 1; i <= 6; i++){            if(a[i]*i >= sum){                for(int j = i; j <= sum; j++){                    if((j-i > 0 && d[j-i]) || j-i == 0)                        d[j] = 1;                }            }            else if(a[i]){                int k = a[i]/2;                k /= 2;                int j;                for(j = 1; j <= k; j=(j<<1)){                    for(int z = sum; z >= j*i; z--){                        if((z-i*j > 0 && d[z-i*j]) || z-i*j == 0)                            d[z] = 1;                    }                }                int x = a[i]+1-j;                for(int z = sum; z >= i*x; z--){                    if((z-i*x > 0 && d[z-i*x]) || z-i*x == 0)                        d[z] = 1;                }            }        }        if(d[sum])            printf("Can be divided.\n");        else            printf("Can't be divided.\n");        printf("\n");    }    return 0;}

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