HDOJ2033 && HDOJ2034

★本题目考察代码熟练度，没有用到数据结构和算法

★代码实现

#include<stdio.h>

int main()
{
    int n;
    int AH,AM,AS,BH,BM,BS,SH,SM,SS;
    
    scanf("%d",&n);
    while(n --)
    {
        SH = SM = SS = 0;
        scanf("%d %d %d %d %d %d",&AH,&AM,&AS,&BH,&BM,&BS);                          
        
        SS = AS + BS;
        if(SS >= 60)
        {
            SS -= 60;
            SM += 1;
        }     
        
        SM = SM + AM + BM;
        if(SM >= 60)  
        {
            SM -= 60;
            SH += 1;      
        }          
        SH = SH + AH + BH;
        printf("%d %d %d\n",SH,SM,SS); 
    }       
    return 0;    
} 

★本题比较坑，输出格式搞了好久

★代码实现

#include <stdio.h>

int main()
{
    int m,n,i,j,flag,count,temp;
    int a[250],b[105];
    while(scanf("%d %d",&m,&n)!=EOF && (m!=0 || n!=0))
    {
        count = 0;
        for(i = 0;i < m+n;i++)
            scanf("%d",&a[i]);   
         
        for(i = 0;i < m;i++)
        {
            flag = 0;
            for(j = m;j < m+n;j++)
            {
                if(a[i]==a[j])
                {
                    flag = 1;  
                    break;                
                }
            }
            
            if(flag == 0)
            {
                b[count] = a[i];      
                count++; 
            }      
        } 
        
        
        for(i = 0;i < count;i++)
        {
            for(j = 0;j < count-i-1;j++)
            {
                if(b[j] > b[j+1])
                {
                    temp = b[j];
                    b[j] = b[j+1];
                    b[j+1] = temp;                
                }              
            }      
        } 
        
        if(count == 0)
            printf("NULL\n");
        else
        {    
            for(i = 0;i < count;i++)
                printf("%d ",b[i]);
            printf("\n");      
        } 
    }
    return 0;    
}

HDOJ2033 && HDOJ2034

原文：http://blog.csdn.net/xgsilence/article/details/40587823