题目描述

小D是某钢材公司的董事长，公司与外国多家企业长期有良好的合作关系；这天，他们公司要运一批货到国外；于是，他指定他的副总小Q到XX码头将这批货物运往A国，到了码头，小Q犯难了；船所能载的重量只有W；如何才能把更多的集装箱运出去呢？

输入格式

输入数据有多组，每组的第一行W代表船的载重量；第二行N代表N各集装箱；第三行是每个集装箱的重量

输出

输出能带走的集装箱的个数

样例输入

5
3
1 2 3
样例输出

2



#include <stdio.h>
int main()
{
    int shipWeight,shipCount;
    scanf("%d",&shipWeight);
    scanf("%d",&shipCount);
   
    int weight[shipCount];
    for(int i = 0;i < shipCount;i++)
        scanf("%d",&weight[i]);
   
    int temp;
    for(int j = 0;j < (shipCount-1);j++)
    {
        for(int k = 1;k < shipCount;k++)
        {
            if(weight[k]<weight[k-1])
            {
                temp = weight[k];
                weight[k] = weight[k-1];
                weight[k-1] = temp;
            }
        }
    }   
   
    int m,total = 0,count = 0;
    for(m = 0;m < shipCount;m++)
    {
        total = total + weight[m];
        if(total <= shipCount)
            count++;   
    }
    printf("%d",count);
    return 0;
}

#include <stdio.h>
int main()
{
    int shipWeight,shipCount,i;
    scanf("%d",&shipWeight);
    scanf("%d",&shipCount);
    
    int weight[shipCount];
    for(i = 0;i < shipCount;i++)
        scanf("%d",&weight[i]);
    
    int temp,j,k;
    for( j = 0;j < (shipCount-1);j++)
    {
        for( k = 1;k < shipCount;k++)
        {
            if(weight[k]<weight[k-1])
            {
                temp = weight[k];
                weight[k] = weight[k-1];
                weight[k-1] = temp;
            }
        }
    }    
    
    int m,total = 0,count = 0;
    for(m = 0;m < shipCount;m++)
    {
        total = total + weight[m];
        count++;//这里有点问题，因为你是要把那些重量用完， 
        
        if(shipWeight-total<weight[m+1])//所以应该是重量来作为判断 
            break;    
    }
    printf("%d",count);
    return 0;
}

说一下判断的思路，就是就是看剩下的重量还能不能够装下一个最轻的货物，