URAL 1047 Simple Calculations

URAL_1047

根据原式可以得a[n+1]-a[n]=a[n]-a[n-1]+2*c[n]①，设S[n]=c[1]+c[2]+…+c[n]，对①式叠加相消可以得到a[n+1]-a[1]=a[n]-a[0]+2*S[n]，整理一下得到a[n+1]-a[n]=a[1]-a[0]+2*S[n]②，对②式叠加相消可以得到a[n+1]-a[1]=n*(a[1]-a[0])+2*(S[1]+S[2]+…+S[n])，这样就可以求得a[1]了。

#include<stdio.h>
#include<string.h>
int N;
void solve()
{int i;double x, y, s = 0, a = 0, c;scanf("%lf%lf", &x, &y);for(i = 0; i < N; i ++){scanf("%lf", &c);a += c;s += a;}printf("%.2f\n", -(2 * s - N * x - y) / (N + 1));
}
int main()
{while(scanf("%d", &N) == 1){solve();}return 0;
}

转载于:https://www.cnblogs.com/staginner/archive/2012/05/07/2487170.html

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URAL 1047 Simple Calculations

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