标签:大数 tin width pop include sizeof each main text
/*A + B Problem II
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 = 1111111111111111110*/
/*借鉴大神经验。与新手共勉。
*/
#include<stdio.h> #include<string.h> int main() { int str1[1100],str2[1100]; char a[1100],b[1100]; int test,i,j,t,k=1,len1,len2; scanf("%d",&test); getchar(); while(test--) { char sum[1100]={0}; scanf("%s %s",a,b); getchar(); len1=strlen(a); len2=strlen(b); memset(str1,0,sizeof(str1)); memset(str2,0,sizeof(str2)); for(i=0,j=len1-1;i<len1;i++,j--) str1[j]=a[i]-'0'; for(i=0,j=len2-1;i<len2;i++,j--) str2[j]=b[i]-'0'; if(len1<len2) { t=len2; len2=len1; len1=t; } for(i=0;i<=len1;i++) { sum[i]=str1[i]+str2[i]+sum[i]; if(sum[i]/10>0) { sum[i+1]=sum[i]/10; sum[i]=sum[i]%10; } } printf("Case %d:\n",k); k++; printf("%s + %s = ",a,b); if(sum[len1]!=0) printf("%d",sum[len1]); for(j=len1-1;j>=0;j--) printf("%d",sum[j]); printf("\n"); if(test!=0) printf("\n"); } return 0; }
今天尝试用java写这道大数题目,贴下代码:
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
int test=in.nextInt();
String s1,s2;
int kase=1;
while(test-->0)
{
s1=in.next();
s2=in.next();
BigInteger s3=new BigInteger(s1);
BigInteger s4=new BigInteger(s2);
System.out.println("Case "+kase+++":");
System.out.println(s3+" + "+s4+" = "+s3.add(s4));
if(test!=0)
System.out.println();
}
}
}
标签:大数 tin width pop include sizeof each main text
原文地址:http://www.cnblogs.com/wzjhoutai/p/7152751.html