杭电 1002 A + B Problem II(大数处理)
发布时间:2020-12-14 03:04:37 所属栏目:大数据 来源:网络整理
导读:A + B Problem II Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 209179????Accepted Submission(s): 40226 Problem Description I have a very simple problem for you. Given two integers
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 209179????Accepted Submission(s): 40226
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
?
Sample Output
?
Author
Ignatius.L
?
?
就是一简单的大数,可是因为后面的换行符没有考虑周全,PE了六回
?
代码如下:
#include<stdio.h> #include<string.h> int main() { int n,i,j,t,p,q,d,k; char a[1010],b[1010]; int c[1010]; scanf("%d",&n); //getchar(); for(t=1;t<=n;t++) { scanf("%s%s",a,b); p=strlen(a); q=strlen(b); d=0; for(i=p-1,j=q-1,k=0;i>=0&&j>=0;i--,j--,k++) { d+=a[i]-'0'+b[j]-'0'; c[k]=d%10; d/=10; } if(k==p) { while(j>=0) { d+=b[j]-'0'; c[k]=d%10; d/=10; j--;k++; } } else { while(i>=0) { d+=a[i]-'0'; c[k]=d%10; d/=10; i--;k++; } } if(d!=0) { c[k]=d; k++; } printf("Case %d:n",t); printf("%s + %s = ",b); for(i=k-1;i>=0;i--) printf("%d",c[i]); printf("n"); if(t!=n)//此处就是控制的条件 printf("n"); } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |