题目来源:http://acm.hdu.edu.cn/showproblem.php?pid=1071
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s):
7011 Accepted Submission(s):
4934
#include <cstdlib> #include <cstring> #include <algorithm> #include <cstdio> #include <cmath> #include <iostream> #include <vector> using namespace std; typedef long long ll; double x1,y11,k,b,a; // 积分 double f(double x) // 不定积分值 { return a*(x-x1)*(x-x1)*(x-x1)/3.0 - k/2*x*x+(y11-b)*x; } int main() { double x2,y2,x3,y3; int t; cin>>t; while(t--) { cin>>x1>>y11>>x2>>y2>>x3>>y3; a=(y3-y11)/((x3-x1)*(x3-x1)); k=(y3-y2)/(x3-x2); b=y3-k*x3; printf("%.2f\n",f(x3)-f(x2)); // 定积分代入上下限 } return 0; }
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原文:http://www.cnblogs.com/zn505119020/p/3616776.html