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 = 1111111111111111110
题解:模拟加法,加进位之间的关系即可
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main () {
char a[8000],b[8000];
int na[8000],nb[8000],sum[8000],pre,flag=1;
int t;
scanf("%d",&t);
while (t--) {
memset(sum,0,sizeof(sum));
memset(na,0,sizeof(na));
memset(nb,0,sizeof(nb));
scanf("%s%s",a,b);
pre=0;
int lena=strlen(a);
int lenb=strlen(b);
for (int i=0; i<lena; i++)
na[lena-1-i]=a[i]-'0';
for (int j=0; j<lenb; j++)
nb[lenb-1-j]=b[j]-'0';
int lenx=lena>lenb?lena:lenb;
for (int k=0; k<lenx; k++) {
sum[k]=na[k]+nb[k]+pre/10;
pre=sum[k];
}
while (pre>9) {
sum[lenx]=pre/10%10;
lenx++;
pre/=10;
}
printf ("Case %d:
",flag++);
printf ("%s + %s = ",a,b);
for (int i=lenx-1; i>=0; i--) {
printf ("%d",sum[i]%10);
}
printf ("
");
if (t)
printf ("
");
}
return 0;
}