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  • hdu1002

    A + B Problem II

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 141857    Accepted Submission(s): 26917

    Problem Description
    I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
     
    Input
    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.
     
    Output
    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. Output a blank line between two test cases.
     
    Sample Input
    2 1 2 112233445566778899 998877665544332211
     
    Sample Output
    Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
     
     
     
     
    //此题为高精度,不可大意
    #include<stdio.h>
    #include<string.h>
    int main()
    {
     char op1[1002],op2[1002];
     int c,i,j,n,m,s1[1002],s2[1002],len1,len2,count;
     count=1;
     scanf("%d",&n);
     m=n;
     while(m--)
     {
      memset(s1,0,1002*sizeof(int));
      memset(s2,0,1002*sizeof(int));
      scanf("%s",op1);
      scanf("%s",op2);
      len1=strlen(op1);
      len2=strlen(op2);
      c=0;
      for(i=len1-1;i>=0;i--)
       s1[c++]=op1[i]-'0';//把顺序颠倒
      c=0;
      for(i=len2-1;i>=0;i--)
       s2[c++]=op2[i]-'0';
      for(i=0;i<1002;i++)
      {
       s1[i]+=s2[i];
       if(s1[i]>=10)
       {
        s1[i]-=10;//逢十进位
        s1[i+1]++;
       }
      }
      printf("Case %d:\n",count++);
      printf("%s + %s = ",op1,op2);
      for(i=1001;i>=0;i--)
       if(s1[i])//清掉前导0
        break;
      for(j=i;j>=0;j--)
       printf("%d",s1[j]);
       printf("\n");
     }
      printf("\n");
      if(count!=n+1)
     return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/Deng1185246160/p/2940402.html
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