UVa 10014 Simple calculations

題目連結
想法：
a[i] = (a[i-1]+a[i+1])/2-c[i]，代入i値，左右兩式從1加到n，消除多餘項目後可得到：

• a[1]+a[n] = a[0]+a[n+1] - 2*(c[1]+...+c[n])

再從這個等式代入n値，左右兩式從1加到n，可得到

• (n+1)*a[1] = n*a[0] + a[n+1] - 2*[n*c[1] + (n-1)*c[2] + ... + 1*c[n]]

此時只剩下a[1]為未知數。

	#include <cstdio>
	#include <algorithm>
	using namespace std;
	double a0,an1,c[3010];
	int Case, N;
	double sum_c()
	{
	double sum = 0;
	for (int i = 1, j = N; i <= N; i++, j--)
	sum += (c[i] * j);
	return sum;
	}
	int main()
	{
	scanf("%d", &Case);
	while (Case--){
	scanf("%d", &N);
	scanf("%lf %lf", &a0, &an1);
	for (int i = 1; i <= N; i++)
	scanf("%lf",&c[i]);
	printf("%.2f\n", (N * a0 + an1 - 2 * sum_c()) / (N+1));
	if (Case) printf("\n");
	}
	return 0;
	}

view raw UVa 10014 Simple calculations.cpp hosted with ❤ by GitHub