Arithmetic Expression
 
时间限制:2000ms
单点时限:200ms
内存限制:256MB

描述

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

输入

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.

输出

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

<dl class="des"> <dt> 样例输入 </dt> <dd> 
4
901 / 100
3 * 3
2 + 6
8 - -1
</dd> <dt> 样例输出 </dt> <dd> 
2

【题意】：算出的结果最接近9的为第几个，相同的话输出靠前的。注意卡精度！要用double~
【代码】：
#include <bits/stdc++.h>
#define LL long long
#define maxn 500005
const int inf = 0x3f3f3f3f;
using namespace std;

int n,idx;
double a,b,sum=0;
char op;
double mi=0x3f3f3f3f3f;
void cal()
{
    switch(op)
    {
    case '+':
        sum=a+b;
        break;
    case '-':
        sum=a-b;
        break;
    case '*':
        sum=a*b;
        break;
    case '/':
        sum=a/b;
        break;

    }
}
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a>>op>>b;
        cal();
        double now = abs(9-sum);
        if(now < mi)//在线查询，边输入边查询，也可以叫打擂台算法，谁小谁上当min King,并记录是第几个人
        {
            idx=i;//而且是now < mi 不能等于，无形记录了字典序最小的那个人，因为后面出现的也是相等，不是小于了！
            mi=now;
        }
    }
    cout<<idx<<endl;
}
打擂台 严格小于

 

</dd> </dl>