This is an A+B Problem
Time Limit: 1000 ms Memory Limit: 65536 KiB
Problem Description
As usual, there will be an A+B problem in warming up, this problem is:
Given two integers A and B, your job is to calculate the sum of A + B.
Given two integers A and B, your job is to calculate the sum of A + B.
Input
There are several test cases, For each test case:
There are two integers A, B for each case (0 ≤ A , B < 101000).
There are two integers A, B for each case (0 ≤ A , B < 101000).
Output
For each test case, output one line containing the result of A+B.
Sample Input
1 2 11111111111 11111111111
Sample Output
3 22222222222
提示:这个题看似非常简单,其实他说是A+B Problem,但是其实看样例就满满的恶意,10的1000次方,用long long也过不了,
所以这不是一个简单的A+B Problem,所以这道题要用到大数模拟,即用字符串数组模拟来完成两数相加的操作。
代码实现如下(g++):
#include <bits/stdc++.h> using namespace std; char s1[1010]; char s2[1010]; int ab[1010];//因为是1000次方,所以只需开1010的数组即可 int main() { int i,j; int len1,len2,a,b; memset(s1,'\0',sizeof(s1)); memset(s2,'\0',sizeof(s2)); memset(ab,0,sizeof(ab));//重置数组 while(~scanf("%s %s",s1,s2)) { len1=strlen(s1); len2=strlen(s2); if(len1==1&&s1[0]=='0'&&len2==1&&s2[0]=='0')//当出现0+0=0时 { printf("0\n"); continue; } int y=0; for(i=len1-1,j=len2-1; i>=0||j>=0; i--,j--) { if(i>=0) { a=s1[i]-'0'; } else { a=0; } if(j>=0) { b=s2[j]-'0'; } else { b=0; } ab[y++]=a+b; } for(i=0; i<y; i++) { ab[i+1]+=(ab[i]/10); ab[i]%=10;//数组储存每一位 } i=y; while(ab[i]==0) { i--; } for(; i>=0; i--) { printf("%d",ab[i]); } printf("\n"); memset(s1,'\0',sizeof(s1)); memset(s2,'\0',sizeof(s2)); memset(ab,0,sizeof(ab));//必须重置数组 } return 0; } /*************************************************** Result: Accepted Take time: 0ms Take Memory: 156KB ****************************************************/