HDOJ 题目1002A + B Problem II (大数)
发布时间:2020-12-14 03:27:53 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): 209712????Accepted Submission(s): 40343 Problem Description I have a very simple problem for you. Given two integers
A + B Problem IITime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 209712????Accepted Submission(s): 40343
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.
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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.
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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.
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Sample Input
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Sample Output
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Author
Ignatius.L
?题目大意:大数的加法
ac代码
#include<stdio.h> #include<string.h> int main() { int cot=0,n; char a[1010],b[1010]; scanf("%d",&n); while(n--) { int c[1100],len1,len2,i,j,k=0,max; getchar(); scanf("%s %s",a,b); len1=strlen(a); len2=strlen(b); i=len1-1; j=len2-1; memset(c,sizeof(c)); while(i>=0||j>=0) { if(i<0&&j>=0) c[k]+=b[j]-'0'; else if(i>=0&&j<0) c[k]+=a[i]-'0'; else c[k]+=(a[i]-'0')+(b[j]-'0'); k++; c[k]=c[k-1]/10; c[k-1]%=10; if(c[k]) max=k; else max=k-1; j--; i--; } printf("Case %d:n",++cot); printf("%s + %s = ",b); for(j=max;j>=0;j--) printf("%d",c[j]); printf("n"); if(n) printf("n"); } } (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! |