POJ2602《Superlong sums》方法：高精度 模拟

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int n; // 输入长度

int main()
{
	//freopen("temp.txt", "r", stdin);
	cin >> n;
	int *a = new int
;
	int *b = new int
;
	int *ans = new int
;
	char *ansc = new char[n+1];
	for (int i = n-1; i >= 0; --i) 
		scanf("%d %d", &a[i], &b[i]);
	
	memset(ans, 0, sizeof(ans) * n);  //当ans为int ans
时，用sizeof是不同的
	int carry = 0;
	for (int i = 0; i < n; ++i) { //这题居然结果与输入结果长度一致
		ans[i] += a[i] + b[i] + carry;
		carry = ans[i] / 10;
		ans[i] = ans[i] % 10;
	}
	for (int i = 0; i < n; ++i)   // 这里不能直接循环输出ans，会超时
		ansc[i] = ans[n-1-i] + '0';
	ansc
 = '\0';
	cout << ansc << endl;

	return 0;
}