1151: large integer addition

1151: Large integer addition

Title Description
Billy often encounters the addition of very large integers , And ordinary calculators can't do . So he wants you to help him write a program to calculate the result .
Input
Multiple groups of input data . First, enter an integer T, Express T Group input .
Enter two large integers for each group , And separated by spaces . Each integer has a maximum of 1000 position . No negative number input .
Output
For each group of input , Output the sum of two integers , Take a line alone .
The sample input Copy
2
1 2
112233445566778899 998877665544332211
Sample output Copy
3
1111111111111111110
source / classification

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#define N 1001
/*  Large integer addition , In the form of string   In reverse order , This makes it easy to calculate from individual bits  a,b,c,r Arrays represent addends, respectively , Addition number , Sum and remainder , carry  */

int main(){
    
    int t,m,n,max=0;
    scanf("%d",&t);
    char s1[N],s2[N];
    int a[N],b[N],c[N],r[N];

    while(t--){
    
        scanf("%s%s",&s1,&s2);
        m=strlen(s1);
        n=strlen(s2);
        
        // Number of digits of the result <= The bigger one 
        if(m<n) max=n;
        else max=m;
        
        // Set it to... Every time 0
        for(int i=0;i<=max;i++){
    
            a[i]=b[i]=c[i]=r[i]=0;
        }
        
        // Reverse the string and change the character type to int
        for(int i=0;i<m;i++){
    
            a[m-i-1]=s1[i]-'0';
        }
        for(int i=0;i<n;i++){
    
            b[n-i-1]=s2[i]-'0';
        }
        
        
        for(int i=0;i<=max;i++){
    
        	// direct +c[i], Find the remainder , And keep it in r[i]
            r[i]=(a[i]+b[i]+c[i])%10;
            // If there is carry ,c[i+1] It's from 0 Change to carry number 
            if((a[i]+b[i]+c[i])/10!=0) c[i+1]=(a[i]+b[i]+c[i])/10;
        }
        
        // The last bit does not necessarily have a carry , namely c[max] It could be 0
        if(r[max]!=0) printf("%d",r[max]);
        for(int i=max-1;i>=0;i--){
    
            printf("%d",r[i]);
        }
        printf("\n");
    }
    return 0;
}