UVA - 10014 Simple calculations
2016-07-19 21:23
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题目大意:根据所给公式和已知量求 a1
解题思路:由所给公式化 a1 的公式。
侵删。 uva 10014 - Simple calculations by zhaofukai
从第一个式子往下累加,整理,得到 n 个式子:
n个式子相加,得:
即 a1 表达式
解题思路:由所给公式化 a1 的公式。
侵删。 uva 10014 - Simple calculations by zhaofukai
2ai = ai-1 + ai+1 - 2ci |
---|
2a1 = a0 + a2 - 2c1 |
2a2 = a1 + a3 - 2c2 |
2a3 = a2 + a4 - 2c3 |
…… |
2an = an-1 + an+1 - 2cn |
a1 + a1 = a0 + a 2 - 2c1 |
---|
a1 + a2 = a0 + a3 - 2(c1 + c2) |
a1 + a3 = a0 + a4 - 2(c1 + c2 + c3) |
…… |
a1 + an = a0 + a - 2(c1 + c2 + c3 + …… + cn) |
(n+1)a1 = na0 + an+1 - 2(nc1 + (n-1)c2 + (n-2)c3 + …… + cn) |
---|
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> double C[10000]; using namespace std; int main() { int T; double a1, a0, a; scanf("%d", &T); while(T--) { int n; scanf("%d", &n); scanf("%lf%lf", &a0, &a); for (int i = 1; i <= n; i++) scanf("%lf", &C[i]); double sum = 0; for (int i = 1; i <= n; i++) sum += (n+1-i) * C[i]; a1 = (n * a0 + a - 2 * sum) / (n + 1); printf("%.2lf\n", a1); if(T) printf("\n"); } return 0; }
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