SCUT Training 20170913 Problem B
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原题:http://acm.hdu.edu.cn/showproblem.php?pid=1012
思路:
水题,注意一开始的输出和n==0、n==1、n==2时小数点位数。每次计算求和后输出即可。
源代码:
#include<cstdio>int main(){ int n,b; double e=2.00,a=1.00; printf("n e\n"); printf("- -----------\n"); for(n=0;n<=9;n++) { printf("%d ",n); if(n==0) { b=1; printf("%d\n",b); continue; } if(n==1) { b=2; printf("%d\n",b); continue; } if(n>=2) { a=a/n; e=e+a; } if(n==2) printf("%.1lf\n",e); else printf("%.9lf\n",e); } return 0;}
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