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.
- 样例输入
-
4 901 / 100 3 * 3 2 + 6 8 - -1
- 样例输出
-
2
程序：
1 #include<iostream> 2 #include<string> 3 #include<stdlib.h> 4 #include<math.h> 5 using namespace std; 6 7 double caluculate(double a, double b, char op) 8 { 9 if(op == '+') 10 { 11 return a + b; 12 } 13 else if(op == '-') 14 { 15 return a - b; 16 } 17 else if(op == '*') 18 { 19 return (double)(a * b); 20 } 21 else if(op == '/') 22 { 23 return (double)(a / b); 24 } 25 } 26 27 int main(void) 28 { 29 int amount = 0; 30 cin>>amount; 31 double tmpmin = 2147483647; 32 int position = -1; 33 for(int i=0; i<amount; ++i) 34 { 35 double a,b; 36 char op; 37 cin>>a; 38 cin>>op; 39 cin>>b; 40 double tmpresult = caluculate(a,b,op); 41 double tmp_result = fabs(9-tmpresult); 42 if(tmp_result < tmpmin) 43 { 44 tmpmin = tmp_result; 45 position = i+1; 46 } 47 } 48 49 cout<<position<<endl; 50 51 return 0; 52 }
遇到的问题：
- 一开始没有发现结果应该是double的，int型的话会导致除法的小数位看不到
- 刚开始使用了stdlib.h中的abs()函数，后来发现应该用math.h中的fabs(),看来自己对math.h中的许多函数都不是很熟悉，不能熟练应用，随后补充
