A1 = ? （数学归纳法）

来源：互联网发布：乐视视频无法连接网络编辑：程序博客网时间：2024/05/29 08:14

A1 = ?

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4294 Accepted Submission(s): 2751

Problem Description

有如下方程：A_i = (A_i-1 + A_i+1)/2 - C_i (i = 1, 2, 3, .... n).
若给出A₀, A_n+1, 和 C₁, C₂, .....C_n.
请编程计算A₁ = ?

Input

输入包括多个测试实例。
对于每个实例，首先是一个正整数n,(n <= 3000); 然后是2个数a₀, a_n+1.接下来的n行每行有一个数c_i(i = 1, ....n);输入以文件结束符结束。

Output

对于每个测试实例，用一行输出所求得的a1(保留2位小数).

Sample Input

150.0025.0010.00250.0025.0010.0020.00

Sample Output

27.5015.00

已知A1=(A0+A2)/2 - C1, A2=(A1+A3)/2 - C2 , ...=>A1+A2 = (A0+A2+A1+A3)/2 - (C1+C2)=>A1+A2 = A0+A3 - 2(C1+C2) 类似的有：A1+A2 = A0+A3 - 2(C1+C2)A1+A3 = A0+A4 - 2(C1+C2+C3)A1+A4 = A0+A5 - 2(C1+C2+C3+C4)...A1+An = A0+An+1 - 2(C1+C2+...+Cn)A1+A1 = A0+A2 - 2(C1) (本来就是)----------------------------------------------------- 左右求和(n+1)A1+(A2+A3+...+An) = nA0 +(A2+A3+...+An) + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)=> (n+1)A1=nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)=> A1 = [nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)]/(n+1)

#include<iostream>#include<string.h>#include<iomanip>using namespace std;#define N 3002double a[N],c[N],sum[N];int main(){int n;while(cin>>n){int i;double sum1=0;//memset(a,0,sizeof(a));cin>>a[0]>>a[n];memset(sum,0,sizeof(sum));//memset(c,0,sizeof(c));for(i=1;i<=n;++i){cin>>c[i];c[i]=2*c[i];sum[i]=sum[i-1]+c[i];sum1+=sum[i];}cout<<setiosflags(ios::fixed)<<setprecision(2)<<(n*a[0]+a[n]-sum1)/(n+1)<<endl;}return 0;}

0 0