UVA10014- Simple calculations

思路：剛開始依照題目所給的a[i]= (a[i - 1] + a[i + 1]) / 2 - c[i]，列出1到i的式子，然後全部相加，化簡，得到a[1] = a[i + 1] - a[i] + a[0] - 2C(i)；

            之後就蒙了。。。不懂怎麼搞了，後來網上看來下才發現化簡的式子在進行累加的話，可以把a[i]化去，最後得到公式（i + 1）a[1] = a[i + 1] + n * a[0] - 2 * (C(1)-C(i))；

#include<stdio.h>#include<string.h>#define N 3005int main() {int cas;double c[N], sum, a, b;scanf("%d", &cas);while (cas--) {int n, m;scanf("%d", &n);scanf("%lf %lf", &a, &b);memset(c, 0, sizeof(c));for(int i = 1; i <= n; i++)scanf("%lf", &c[i]);sum = 0;m = n;for(int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++) sum += c[j];m--;}printf("%.2lf\n", (b + n * a - 2 * sum) / (n + 1));if (cas)printf("\n");}return 0;}