题目挺简单，主要是思想不难。但很多小点要注意。

题目：子序列的和(subsequence)

输入两个整数n<m<10^6,输出1/(n^2) + 1/((n+1)^2) + 1/((n+2)^2) 1/((n+3)^2) + ... + 1/((m)^2)，保留5位小数。输入包含多组数据，结束标记为n=m=0。提示：本题目有陷阱。

样例输入：
2 4
65536 655360

0 0

样例输出：

Case 1: 0.42361

Case 2: 0.00001

 1 #include <stdio.h>
 2 #include <math.h>
 3 
 4 int main () {
 5     int n, m;
 6     int p = 1;
 7     while (scanf ("%d%d", &n, &m) != EOF) { 
 8         if (n == 0 && m == 0) 
 9             break;
10         double sum = 0;
11         for (int i = n; i <= m; i++) {
12             sum += 1.0 / (long long)(pow(i, 2));
13         }
14         printf ("Case %d: %.5f\n", p, sum);
15         p++;
16     }
17     return 0;
18 }

 1、 多组数据

while(cin >> m >> n >> t>>s){}
while (~scanf ("%d%d%d%d", &m, &n, &t, &s)) {}
while (scanf ("%d", &a) != EOF) {}
while (scanf ("%d", &a) == 1) {}

2、 continue与break

continue 直接进行下一次循环
break直接跳出循环

3、 小心乘法溢出

4、 平方开根绝对值（#include <math.h>） 

几次方 pow(，)
int a = pow(4,2);4^2
int a = pow(4,0.5);4^0.5
开方
int b = sqrt(4);
整数绝对值
int a = abs(b-c);
浮点数绝对值
double a = fabs(b-c);