PAT乙题1010. 一元多项式求导 (25)

2018/01/25 PAT 共 660 字,约 2 分钟
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<map>
#include<string>
#include<fstream>
#include<vector>
#include<stack>
#include<functional>
using namespace std;

int main() {
#ifdef LOCAL
	freopen("d:\\datain.txt", "r", stdin);
	freopen("d:\\dataout.txt", "w", stdout);
#endif LOCAL
	int a[1000];
	int b[1000];
	int i = 0;
	while (scanf("%d%d", &a[i], &b[i])==2) {
		if (b[i] == 0)
			continue;
		a[i] = a[i] * b[i];
		if (b[i] == 0)
			b[i] == 0;
		else
			b[i] = b[i] - 1;
		i++;
	}
	for (int j = 0; j < i-1; j++) {
		cout << a[j] << " " << b[j] << " ";
	}
	cout << a[i - 1] << " " << b[i - 1];
	return 0;
}

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