HDU

发布时间：2025-11-18 23:36

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A1 = ?

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3244 Accepted Submission(s): 2054

Problem Description 有如下方程：Ai = (Ai-1 + Ai+1)/2 - Ci (i = 1, 2, 3, .... n).
若给出A0, An+1, 和 C1, C2, .....Cn.
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, an+1.接下来的n行每行有一个数ci(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 1 50.00 25.00 10.00 2 50.00 25.00 10.00 20.00

Sample Output 27.50 15.00 /* 递推题：逆着写几项就总结出规律了我推出的公式为：a[1]=1/(n+1)*(a[n+1]+n*a[0]-2*c[n]-4*c[n-1]-8*c[n-2]...-pow(2,n)*c[1]); */ //代码一： #include<stdio.h>int main() {int n,i,k;float a[3005],c[3005];float sum;while(~scanf("%d",&n)){scanf("%f%f",&a[0],&a[n+1]);for(i=1;i<=n;++i)scanf("%f",&c[i]);k=2;sum=0;for(i=n;i>0;--i){c[i]*=k;k+=2;sum+=c[i];}printf("%.2f\n",1.0/(n+1)*(a[n+1]+n*a[0]-sum));}return 0; }//代码二：---参考网上的代码 /* 因为：Ai=(Ai-1+Ai+1)/2 - Ci, 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) 同理可得：A1+A1 = A0+A2 - 2(C1) 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) ----------------------------------------------------- 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<stdio.h> int main() {int t,i,n;double a,b,c,a1,sum1,sum2;while(scanf("%d",&n)!=EOF){t=n;scanf("%lf%lf",&a,&b);sum1=t*a+b;sum2=0;for(i=1;i<=n;i++){scanf("%lf",&c);sum2=sum2+t*c;t--;}a1=(sum1-2*sum2)/(n+1);printf("%.2f\n",a1);}return 0; }

转载于:https://www.cnblogs.com/dongsheng/archive/2012/09/13/2684102.html