国外网站怎样建设,广州网站建设南宁,wordpress 4.6 注入,微信营销方式哎 真的是懒得动脑子还是怎么滴。。。 题目如下 Problem Description有如下方程#xff1a;Ai (Ai-1 Ai1)/2 - Ci (i 1, 2, 3, .... n).若给出A0, An1, 和 C1, C2, .....Cn.请编程计算A1 ?参考网上题解。。。因为#xff1a;Ai(Ai-1Ai1)/2 - Ci, A1(A0 A2 )/2 - C1;A2…哎 真的是懒得动脑子还是怎么滴。。。 题目如下 Problem Description 有如下方程Ai (Ai-1 Ai1)/2 - Ci (i 1, 2, 3, .... n).若给出A0, An1, 和 C1, C2, .....Cn.请编程计算A1 ? 参考网上题解。。。 因为Ai(Ai-1Ai1)/2 - Ci, A1(A0 A2 )/2 - C1; A2(A1 A3)/2 - C2 , ... A1A2 (A0A2A1A3)/2 - (C1C2) A1A2 A0A3 - 2(C1C2) 同理可得 A1A1 A0A2 - 2(C1) A1A2 A0A3 - 2(C1C2) A1A3 A0A4 - 2(C1C2C3) A1A4 A0A5 - 2(C1C2C3C4) ... A1An A0An1 - 2(C1C2...Cn) ----------------------------------------------------- 左右求和 (n1)A1(A2A3...An) nA0 (A2A3...An) An1 - 2(nC1(n-1)C2...2Cn-1Cn) (n1)A1 nA0 An1 - 2(nC1(n-1)C2...2Cn-1Cn) A1 [nA0 An1 - 2(nC1(n-1)C2...2Cn-1Cn)]/(n1) #includestdio.h
#includestring.h
#includevector
#includecmath
using namespace std;
int n;double a,b,c;
int main()
{while(scanf(%d,n)!EOF){double ans0;scanf(%lf%lf,a,b);ansn*ab;for(int in;i1;i--){scanf(%lf,c);ans-2*i*c;}printf(%.2f\n,ans/(n1));}return 0;
} AC代码转载于:https://www.cnblogs.com/Geek-xiyang/p/5356778.html
