• A+B Problem II


    A+B Problem II
    时间限制:3000 ms | 内存限制:65535 KB
    难度:3
    描述
    I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

    A,B must be positive.

    输入
    The first line of the input contains an integer T(1<=T<=20) which means the number of test cases.
    Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers
    are very large, that means you should not process them by using 32-bit integer. You may assume the
    length of each integer will not exceed 1000.
    输出
    For each test case, you should output two lines. The first line is "Case #:", # means the number of
    the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note
    there are some spaces int the equation.
    样例输入
    2
    1 2
    112233445566778899 998877665544332211
    样例输出
    Case 1:
    1 + 2 = 3
    Case 2:
    112233445566778899 + 998877665544332211 = 1111111111111111110
    */

    #include<stdio.h>
    #include<string.h>
      int main(){
          int n;
          int len1,len2,i,j;
          int t; 
            int k=0;
           scanf("%d",&n);
           while(n--){
               char a1[1100],a2[1100];// 开始时,数组定义过小,导致超时 
               int a3[1100],a4[1100];
                int b2[1100];
                memset(a3,0,sizeof(a3));
                memset(a4,0,sizeof(a4));
                 memset(b2,0,sizeof(b2));
                scanf("%s%s",a1,a2);
                 k++;
            printf("Case %d:
    ",k);
            printf("%s",a1);
            printf(" + ");
            printf("%s",a2);
            printf(" = ");
               len1=strlen(a1);
               len2=strlen(a2);     //将两个字符数倒序,底位在前,高位在后       
            for(i=0;i<len1;i++){
              a3[i]=a1[i]-'0';
            }
              for(j=0;j<len2;j++){
                  a4[j]=a2[j]-'0';
              }
             for(i=0;i<(len1/2);i++)
                 {
              t=a3[i];
                   a3[i]=a3[len1-i-1];
                       a3[len1-i-1]=t;    }
             
             for(j=0;j<(len2/2);j++){
                  t=a4[j];
                   a4[j]=a4[len2-j-1];
                     a4[len2-j-1]=t;
             }
             
               if(len1<len2)
                 len1=len2;
                 
               
           for(i=0;i<=len1;i++)  // 进位,相加大于9时,向下一位进1,并且将本位去十 
           { 
               b2[i]=(a3[i])+(a4[i]);
                 if(b2[i]>9){
                
                a3[i+1]++; 
              b2[i]-=10;
                        }  }
            
            int i=len1+2;
            while(b2[i]==0)i--;//将高位的零去掉,并且倒序一次输出 
            for(;i>0;i--){
                printf("%d",b2[i]);
                      }
                 
               printf("%d
    ",b2[0]);
               
                if(n)   printf("
    ");
           } 
           
      return 0;
    }
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  • 原文地址:https://www.cnblogs.com/acmgym/p/3691864.html
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