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Problem Description
多项式的描述如下：
1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + …
现在请你求出该多项式的前n项的和。
Input
输入数据由2行组成，首先是一个正整数m（m<100），表示测试实例的个数，第二行包含m个正整数，对于每一个整数(不妨设为n,n<1000），求该多项式的前n项的和。
Output
对于每个测试实例n，要求输出多项式前n项的和。每个测试实例的输出占一行，结果保留2位小数。
Sample Input
2
1 2
Sample Output
1.00
0.50
注意：输出要符合规则
#include<stdio.h> int main() { int n,i,j,a[1000]; float s,t; scanf("%d",&n);
for(i=0;i<n;i++) scanf("%d",&a[i]); for(i=0;i<n;i++) {t=1.0;s=0.0; for(j=1;j<=a[
i];j++) { s+=1.0*t/j; t=-t; } printf("%.2f\n",s); } return 0; }
