算法设计与分析之不定期更新的日常3+贪心

算法设计与分析之不定期更新的日常3+贪心

3.装箱问题

代码实现：

#include <iostream>#include <cstdio>///贪心算法求解装箱问题，分类讨论using namespace std;int countain1[4]={0,7,6,5};int countain2[4]={0,5,3,1};int nTotal;int main(){    int b1,b2,b3,b4,b5,b6;    int c1,c2;    while(scanf("%d%d%d%d%d%d",&b1,&b2,&b3,&b4,&b5,&b6)!=EOF)    {        if(b1==0&&b2==0&&b3==0&&b4==0&&b5==0&&b6==0)            break;        nTotal=b6+b5+b4+(b3+3)/4;        c1=c2=0;        c1=b5*11;        c1+=countain1[b3%4];        c2=b4*5;        c2+=countain2[b3%4];        if(b2<=c2)        {            c1+=(c2-b2)*4;        }        else        {            nTotal+=(b2-c2)/9;            int r2=(b2-c2)%9;            if(r2)            {                nTotal++;                c1+=36-r2*4;            }        }        if(b1>c1)        {            nTotal+=(b1-c1+35)/36;        }        printf("%d\n",nTotal);    }    return 0;}/*1 2 3 4 5 6*/