ACM訓練題

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題目1 : Arithmetic Expression

時間限制:2000ms

單點時限:200ms

內存限制:256MB

描述

Given N arithmetic expressions, can you tell whose result is closest to 9?

輸入

Line 1: N (1 <= N <= 50000).
Line 2..N+1: Each line contains an expression in the format of "a op b" where a, b are integers (-10000 <= a, b <= 10000) and op is one of addition (+), subtraction (-), multiplication (*) and division (/). There is no "divided by zero" expression.

輸出

The index of expression whose result is closest to 9. If there are more than one such expressions, output the smallest index.

樣例輸入

4
901 / 100
3 * 3
2 + 6
8 - -1

樣例輸出

2

//source here
#include<iostream>
using namespace std;

int main(int argc,char **argv)
{
	int num;
	char op;
	long double result=1.79769e+308,tmp;
	int index = 1;
	int num1,num2;
	cin>>num;
	for(int i=0;i<num;i++)
	{
		cin>>num1>>op>>num2;
		switch(op)
		{
		case '+':
						tmp = abs(num1+num2 - 9);
						break;
		case '-':
						tmp = abs(num1-num2 - 9);
						break;
		case '*':
						tmp = abs(num1*num2 - 9);
						break;
		case '/':
						tmp = abs(num1/(double)num2 - 9);
						break;
		default:
			break;
		}
		if (tmp == 0.0)
		{
			index = i+1;
			break;
		}
		if(tmp < result )
		{
			index = i+1;
		}
	}
	printf("%d",index);
}