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1、1．輸入整型數(shù)組求數(shù)組的最小數(shù)和最大數(shù)之和，例如輸入1,2,3,4則輸出為5，當輸入只有一個數(shù)的時候，則最小數(shù)和最大數(shù)都是該數(shù)，例如只輸入1，則輸出為2；另外數(shù)組的長度不超過50參考代碼：#includemain(){????intnum[50]={0};????inti,n;?????printf(“請輸入整型數(shù)組的長度(1~50)：”);????scanf(“%d”,&n);?????????printf(“請輸入整型數(shù)組的元素：”);????for(i=0;i

2、???}?????intmin_num=num[0];????intmax_num=num[0];????for(intj=0;jnum[j])????????????min_num=num[j];????}????intsum=min_num+max_num;????printf(“數(shù)組中最大與最小值之和：%d”,sum);????return0;}??2．求兩個長長整型的數(shù)據(jù)的和并輸出，例如

3、輸入1233333333333333。。。3111111111111111111111111.。。。，則輸出。。。。?#include#include#includemain(){????char*num1,*num2;//兩個長長整型數(shù)據(jù)????char*sum;//????inttemp;intlen_num1,len_num2;//兩個長長整型數(shù)據(jù)的長度????intlen_max,len_min;????num1=(char*)malloc(sizeof(char));????num2=(char*)mal

4、loc(sizeof(char));????printf(“輸入兩個長長整型數(shù)據(jù)：”);????scanf(“%s”,num1);????printf(“輸入兩個長長整型數(shù)據(jù)：”);????scanf(“%s”,num2);????len_num1=strlen(num1);????len_num2=strlen(num2);????len_max=(len_num1>=len_num2)?len_num1:len_num2;????len_min=(len_num1<=len_num2)?len_num1:len_num2;????intlen_max1=len_max

5、;????sum=(char*)malloc(sizeof(char)*len_max);????memset(sum,0×00,len_max+1);//切忌初始化????for(;len_num1>0&&len_num2>0;len_num1–,len_num2–)????{????sum[len_max--]=((num1[len_num1-1]-’0′)+(num2[len_num2-1]-’0′));????}????if(len_num1>0)????{????????sum[len_max--]=num1[len_num1-1]-’0′;????????le

6、n_num1–;????}????if(len_num2>0)????{????????sum[len_max--]=num1[len_num2-1]-’0′;????????len_num2–;????}????for(intj=len_max1;j>=0;j–)//實現(xiàn)進位操作????{????//????temp=sum[j]-’0′;????????if(sum[j]>=10)????????{????????sum[j-1]+=sum[j]/10;????????????sum[j]%=10;????????}????}????char*outsum=(char

7、*)malloc(sizeof(char)*len_max1);????j=0;????while(sum[j]==0)//跳出頭部0元素????????j++;????for(intm=0;m