【基础训练】HDOJ2033 && HDOJ2034

来源：互联网发布：中国企业跨国并购数据编辑：程序博客网时间：2024/04/29 10:48

★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;    }

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