Uva 10014 – Simple calculations
Simple calculations
The Problem
There is a sequence of n+2 elements a0, a1,…, an+1(n <= 3000; -1000 <= ai1000). It is known thatai= (ai–1+ai+1)/2–cifor each i=1, 2, …, n. You are given a0, an+1, c1, … , cn. Write a program which calculates a1.
The Input
The first line is the number of test cases, followed by a blank line.
For each test case, the first line of an input file contains an integer n. The next two lines consist of numbers a0and an+1each having two digits after decimal point, and the next n lines contain numbers ci(also with two digits after decimal point), one number per line.
Each test case will be separated by a single line.
The Output
For each test case, the output file should contain a1in the same format as a0and an+1.
Print a blank line between the outputs for two consecutive test cases.
Sample Input1150.5025.5010.15Sample Output27.85
#include<stdio.h>#include<string.h>int main(){int T, n, i;double first, final, sum, temp, ans;scanf(, &T);while(T–){scanf(, &n);scanf(, &first);scanf(, &final);for(i=1,sum=0; i<=n; ++i){scanf(, &temp);sum += 2.0*(n+1-i)*temp;}ans = n*first+final-sum;printf(, ans/(n+1)););}return 0;}
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