POJ 2601|URAL 1047|Simple Caluculations|数学推导

来源：互联网发布：数据库原理与应用第2版编辑：程序博客网时间：2024/06/14 19:06

http://acm.timus.ru/problem.aspx?space=1&num=1047

题目翻译

有一个N+2个元素的数列：a0,a1,⋯,an+1(1≤N≤3000,−2000≤ai≤2000)，并且满足

a i = 1 2 (a i - 1 + a i + 1) - c i (i = 1, 2, \dots, N)

给定

a0,an+1,c1,⋯,cn，写一个程序计算

a1的值。

输入

输入第一行一个整数N，接下来2行包含a0,an+1，均有2位小数，接下来N行每行一个2位小数ci。

输出

输出a1，2位小数。

样例输入

150.5025.5010.15

样例输出

27.85

题解

累加法：

a 1 a 2 a 3 a n - 1 a n = 1 2 (a 0 + a 2) - c 1 = 1 2 (a 1 + a 3) - c 1 = 1 2 (a 2 + a 4) - c 1 ⋮ = 1 2 (a n - 2 + a n) - c n - 1 = 1 2 (a n - 1 + a n + 1) - c n

每行相加得

a 1 + \dots + a n = 1 2 (a 0 + \dots + a n - 1 + a 2 + \dots + a n + 1) - \sum i = 1 n c i

化简得

a 1 + \dots + a n = a 2 + \dots + a n - 1 + 1 2 (a 0 + a 1 + a n + a n + 1) - \sum i = 1 n c i

移项得

1 2 (a 1 + a n) = 1 2 (a 0 + a n + 1) - \sum i = 1 n c i

乘2得

a 1 + a n = a 0 + a n + 1 - 2 \sum i = 1 n c i

令

n=1,2,3,⋯，有

= = = n a 1 + a 1 + a 2 + \dots + a n (a 1 + a 1) + (a 1 + a 2) + \dots + (a 1 + a n) (a 0 + a 2 - 2 \sum i = 1 1 c i) + (a 0 + a 3 - 2 \sum i = 1 2 c i) + (a 0 + a 4 - 2 \sum i = 1 3 c i) + \dots n a 0 + a 2 + a 3 + \dots + a n + 1 - 2 \sum j = 1 n \sum i = 1 j c i

有

n a 1 + \sum i = 1 n a i = n a 0 + \sum i = 2 n a i + a n + 1 - 2 \sum j = 1 n \sum i = 1 j c i

有

(n + 1) a 1 + \sum i = 1 n a i = n a 0 + \sum i = 1 n a i + a n + 1 - 2 \sum j = 1 n \sum i = 1 j c i

变换为

a1的表达式，程序得。

// POJ 2601, URAL 1047#include <cstdio>int main() {    int n, i;    double a, b, c, ans, sum;    scanf("%d", &n);    scanf("%lf%lf", &a, &b);    sum = 0;    for (i = 0; i < n; ++i) {        scanf("%lf", &c);        sum += (n - i) * c;     }     ans = (n * a + b - 2 * sum) / (n + 1);    printf("%.2lf\n", ans);    return 0;}

阅读全文

0 0