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Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7601 Accepted Submission(s): 3913
#include <cmath> #include <cstdio> double fun(int a, int b) { double ans=sqrt(2.0); int n=a+b; //总结; double sum=0.0; for(int i=1; i<n; i++) { sum=sum+i*1.0*ans; } sum=sum+a*ans; // 相似三角形; for(int i=0; i<n; i++) sum=sum+sqrt(i*i+(i+1)*(i+1)); return sum; } int main() { int t; scanf("%d", &t); while(t--) { double a, b, c, d; scanf("%lf%lf%lf%lf", &a, &b, &c, &d); double e=fun(a, b); double f=fun(c, d); printf("%.3lf\n", fabs(e-f)); } return 0; }
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原文地址:http://www.cnblogs.com/fengshun/p/5271585.html