輸入兩個正整數n<m<10^6,輸出1/n^2+1/(n+1)^2+........+1/m^2,保留5位小數。輸入包含多組資料,結束标記為n=m=0.
樣例輸入:
2 4
65536 655360
0 0
樣例輸出:
case 1: 0.42361
case 2:0.00001
#include <stdio.h>//子序列求和
int main()
{
long long n=1,m=1;
int i=1;
double sum=0.0;
while(n!=0&&m!=0)
{
scanf("%ld%ld",&n,&m);
//printf("%ld%ld",n,m);
while(n<=m)
{
sum=sum+1/(double)(n*n);
//printf("%llf\n",sum);
n=n+1;
}
if(n!=0&&m!=0)
{
printf("case %d: %0.5lf\n",i,sum);
i++;
}
}
}