POJ 1014 Dividing (多重背包or母函数)

母函数求解:

需要注意一点,有个结论,但我还查不到这个结论哪里来的。。。

如果n>=8,可将n改写为12(n为偶数),或者11(为奇数)。

#include<cstdio>
#include<cstring>

using namespace std;

int a[7];

const int maxn=3e7;

int x[maxn];
int y[maxn];

int main(){
    int T=1;
    while(1){
        int ans=0;
        for(int i=1;i<=6;i++){
            scanf("%d",&a[i]);
            if(a[i]>8&&a[i]%2) a[i]=11;
            else if(a[i]>8) a[i]=12;
            ans+=a[i]*i;
        }
        if(!ans) break;
        printf("Collection #%d:\n",T++);
        if(ans%2){
            printf("Can't be divided.\n\n");
        }else{
            for(int i=0;i<=ans/2;i++){
                x[i]=y[i]=0;
            }
            x[0]=1;
            int t=ans/2;
            for(int i=1;i<=6;i++){
                for(int j=0;j<=t;j++){
                    for(int k=0;k<=a[i]&&(j+k*i)<=t;k++){
                        y[j+k*i]+=x[j];
                    }
                }
                for(int j=0;j<=t;j++){
                    x[j]=y[j];
                    y[j]=0;
                }
            }
            if(x[t]){
                printf("Can be divided.\n\n");
            }else{
                printf("Can't be divided.\n\n");
            }
        }
    }

}
/*
3
2 7
7 2
5 5
*/

也可以转化为背包问题:

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

int a[7];

const int maxn=3e7;

int dp[maxn];
int v[maxn];

int main(){
    int T=1;
    while(1){
        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",T++);
        if(sum%2){
            printf("Can't be divided.\n\n");
        }else{

            int f=sum/2;
            int len=0;
            for(int i=1;i<=6;i++){
                int num=a[i];
                int c=1;
                while(num-c>0){
                    num-=c;
                    v[++len]=c*i;
                    c<<=1;
                }
                if(num)
                    v[++len]=num*i;
            }
            for(int i=0;i<=f;i++){
                dp[i]=0;
            }
            for(int i=1;i<=len;i++){
                for(int j=f;j>=v[i];j--){
                    dp[j]=max(dp[j],dp[j-v[i]]+v[i]);
                }
            }
            if(dp[f]==f){
                printf("Can be divided.\n\n");
            }else{
                printf("Can't be divided.\n\n");
            }
        }
    }
    return 0;
}
/*
3
2 7
7 2
5 5
*/

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转载自blog.csdn.net/qq_40679299/article/details/82584523