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程序设计入门4 子序列的和
阅读量:390 次
发布时间:2019-03-05

本文共 517 字,大约阅读时间需要 1 分钟。

子序列的和(subsequence)

输入两个正整数n<m<10^6,输出1/n^2 + 1/(n+1)^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

一,注意事项(关于大整数相乘溢出问题):

 

#include
int main(){ int n; long long int n_2; scanf("%d",&n); n_2 = n * n; printf("%lld", n_2); return 0;}

这里假设n = 111111,则111111*111111为11位的,,,虽然小于long long的64位,,,但是n*n由于两个乘数为int,最后的结果依然会用int(10位)表示,纵然式子的左边是long long,依然会溢出。。。

 

虽然网上说强制转换可以解决问题,即n_2 = (long long)n * n;但我的电脑上的编译器会显示栈溢出。最保险的方法就是把乘数也设成long long.

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