HDOJ 1002 A + B Problem II ( 大数相加)
发布时间：2021-02-18 13:48:05 所属栏目：大数据 来源：网络整理
导读：A + B Problem II（点击进入题目） Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 314071????Accepted Submission(s): 60860 Problem Description I have a very simple problem for you. G
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A + B Problem II（点击进入题目）Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 314071????Accepted Submission(s): 60860 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 = 2222222222222221110? Author Ignatius.L ? Recommend We have carefully selected several similar problems for you:?? 1004? 1005? 1009? 1020? 1010? ? 模拟大数运算后注意输出的格式（英语好就不会是问题，我就是问题）。
代码：
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define MYDD 1103
using namespace std;
int main() {
int long_A,long_B,T;
int sum[MYDD],a[MYDD],b[MYDD];
char A[MYDD],B[MYDD];
scanf("%d",&T);
for(int t=1; t<=T; t++) {
memset(sum,sizeof(sum));
memset(a,sizeof(a));
memset(b,sizeof(b));
scanf("%s %s",A,B);
long_A=strlen(A);
long_B=strlen(B);
int max_l=max(long_A,long_B);
for(int j=long_A-1,i=0; j>=0; j--)
a[i++]=A[j]-'0';
for(int j=long_B-1,i=0; j>=0; j--)
b[i++]=B[j]-'0';
for(int j=0; j<max_l; j++) {
sum[j]=a[j]+b[j]+sum[j];//包证前一位进位的数字要加上
if(sum[j]>9) {
sum[j+1]++;
sum[j]-=10;
}
}
printf("Case %d:n",t);
for(int j=long_A-1; j>=0; j--)
printf("%d",a[j]);
printf(" + ");
for(int j=long_B-1; j>=0; j--)
printf("%d",b[j]);
printf(" = ");
if(sum[max_l]!=0)//首位是否为 0
printf("%d",sum[max_l]);
for(int i=max_l-1; i>=0; i--)
printf("%d",sum[i]);
printf("n");//保证格式正确
if(t!=T)
printf("n");
}
return 0;
}
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