POJ 1503 Integer Inquiry(高精度运算)

//简单高精度运算#include<iostream>using namespace std;struct BigInt//结构体{char data[110];//字符串数据int num[110];//整型数据int len;//位数长度void print()//打印函数{if(num[len] != 0)cout << num[len];//注意长度，得提前判断进位的情况for(int i = len-1;i >= 0;--i)cout << num[i];}};void reverse(BigInt &a)//将数位反转{a.len = strlen(a.data);for(int i = 0;i < a.len;++i)a.num[a.len-1-i] = a.data[i] - '0';//最简易做法}void plus(BigInt &a,BigInt &b){int length = max(a.len,b.len);for(int i = 0;i < length;++i){a.num[i+1] += (a.num[i] + b.num[i])/10;//进位a.num[i] = (a.num[i] + b.num[i])%10;}}int main(){BigInt a,b;memset(a.num,0,sizeof(a.num));memset(b.num,0,sizeof(b.num));cin >> a.data;reverse(a);while(cin >> b.data){if(strlen(b.data) == 1 && b.data[0] == '0')break;reverse(b);plus(a,b);}a.print();cout << endl;return 0;}