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HDU1701 ACMer [浮点数误差]
2015-07-24 06:38:51 来源: 作者: 【 】 浏览:49
Tags:HDU1701 ACMer 点数 误差

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ACMer

Time Limit: 1000/1000 MS ( Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3536 Accepted Submission(s): 1629

Problem Description There are at least P% and at most Q% students of HDU are ACMers, now I want to know how many students HDU have at least?
Input The input contains multiple test cases.
The first line has one integer,represent the number of test cases.
The following N lines each line contains two numbers P and Q(P < Q),which accurate up to 2 decimal places.
Output For each test case, output the minumal number of students in HDU.
Sample Input
1
13.00 14.10

Sample Output
15

这题应该分类到“经典题”还是“烂题”呢?这个真的不好说,WA了2次。精度问题,精度问题。。

AC代码:

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#include 
  
   

int main(){
	int n, m; double a, b;
	scanf("%d", &n);
	while(n--){
		scanf("%lf%lf", &a, &b);
		m = 1;
		while((int)(m * a / 100) >= (int)(m * b / 100)) ++m;
		printf("%d\n", m);
	}
	return 0;
}
  
WA代码:

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#include 
  
   

int main(){
	int n, m; double a, b;
	scanf("%d", &n);
	while(n--){
		scanf("%lf%lf", &a, &b);
		a /= 100; b /= 100;
		m = 1;
		while((int)(m * a) >= (int)(m * b)) ++m;
		printf("%d\n", m);
	}
	return 0;
}
  



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