A+B Problem II 大数加法

A+B Problem II

http://acm.nyist.net/JudgeOnline/problem.php?pid=103

时间限制：3000 ms | 内存限制：65535 KB

难度：3

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

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.

我的代码

#include <stdio.h>
#define MAX 1010
char a[MAX],b[MAX],c[MAX];
int main(int argc, char *argv[])
{
int T,t,d,i,j,x,y,N=0;
scanf("%d",&T);
while(T--){
for(i=0;i<MAX;i++)
a[i]=b[i]=c[i]='0';
//将字符串改为 数字低位在数组的低位
scanf("%s",c);
for(x=0;c[x];x++);
for(i=0;i<x;i++)
a[i]=c[x-1-i];
a[i]='0';

scanf("%s",c);
for(x=0;c[x];x++);
for(i=0;i<x;i++)
b[i]=c[x-1-i];
b[i]='0';
for(i=0;i<MAX;i++)
c[i]='0';
c[i]='0';

//加法运算
d=0; //进位
for(i=0;i<MAX;i++){
t = a[i]+b[i]-2*'0'+d;
if(t>9){
d=1;
c[i]=t%10+'0';
}else{
d=0;
c[i]=t+'0';
}
}

//输出
printf("Case %d:\n",++N);
//考虑 0+0=0
for(i=MAX-1;a[i]=='0'&& i;i--);
for(;i>=0;i--)
printf("%c",a[i]);
printf(" + ");
for(i=MAX-1;b[i]=='0'&& i;i--);
for(;i>=0;i--)
printf("%c",b[i]);
printf(" = ");
for(i=MAX-1;c[i]=='0' && i;i--);
for(;i>=0;i--)
printf("%c",c[i]);
printf("\n");
}
return 0;
}

简单做法

#include <stdio.h>
#include <string.h>
#define MAX 1020
char a[MAX],b[MAX],c[MAX];
int main(int argc, char *argv[])
{
int n=0,T,t,i,j,d,lena,lenb,lenc;
scanf("%d",&T);
while(T--){
scanf("%s",a);
scanf("%s",b);
lena=strlen(a);
lenb=strlen(b);
printf("Case %d:\n",++n);
printf("%s + %s = ",a,b);
d=lenc=0;
i=lena-1;
j=lenb-1;
while(i>=0 && j>=0){
t=a[i--]+b[j--]-2*'0'+d;
if(t>9){
d=1;
c[lenc++]=t%10;
}else{
d=0;
c[lenc++]=t;
}
}
while(i>=0){
t=a[i--]-'0'+d;
if(t>9){
d=1;
c[lenc++]=t%10;
}else{
d=0;
c[lenc++]=t;
}
}

while(j>=0){
t=b[j--]-'0'+d;
if(t>9){
d=1;
c[lenc++]=t%10;
}else{
d=0;
c[lenc++]=t;
}
}
if(d)
c[lenc++]=1;

for(t=lenc-1;t>=0;t--)
printf("%d",c[t]);
printf("\n");
}
return 0;
}

题目推荐

import java.math.BigInteger;
import java.util.Scanner;
public class Main{

public static void main(String args[]) {
Scanner cin=new Scanner(System.in);
int n=cin.nextInt();
BigInteger a,b;
for(int i=1;i<=n;i++){
a=cin.nextBigInteger();
b=cin.nextBigInteger();
System.out.println("Case "+i+":");
System.out
7fe0
.println(a.toString()+" + "+b.toString()+" = "+a.add(b));
}
}
}