Topic:
Now, 11.11 in college life is a special day, many schools have it as a boy's day, as a sister, LPT also want to give the younger brother, but so much, how to send? So she thought of an idea,
Can out a math problem exam, the first three AC this problem of the younger brother can get small gifts!
There is a set of sequences, it has n+2 elements, a[0],a[1], ..., a[n+1] (n <=, -10000 <= a[i]<=10000), for i = 1,2,3,...,n satisfies a[i]-c[i]= (a[i-1]+a [I+1]) /2-b[i],
If given a[0],a[n+1] and the values of all elements of set B, C (i.e. given b[1],b[2]...b[n] and c[1],c[2],..., c[n],-100≤b[i], c[i]≤100), are reserved for the two digits after the decimal point, the Smart
Can you figure out a[1]?
Ideas:
first, constant items can be combined, ci= (ai+1"/2 − with - ci bashi sub-formation: < Span class= ">aaa
variants: a - - = -
Then, i 1 start, to i n
a " a1 2
A3 - a2 - 2d2 = a2 - A 1
···
An+1 - an - 2dn = an - an-1
Add the n to get
a0 -< Span class= "> a1 + n+1n * $ \sum_{i=1}^{n}d_i$< Span class= ">= 0 (
by formula (1) available
An+1 - an - 2dN= an - an-1
2
a0 - 1nn-12 * $ \sum_{i=1}^{n-1}d_i$ = (
Repeat , get:
A0 - a1 + a2 - a1 - 2 d1 = 0 (n+1)
The formula (2) is added to the formula (n+1)to:
A1=(n*A0+AN+1-2* * 1 -2 * 2dn
Code:
1#include <cstdio>2 intMain ()3 {4 //freopen ("Data_in.txt", "R", stdin);5 //freopen ("Data_out.txt", "w", stdout);6 intN, I, T;7 DoubleA1, X, Y, b[3010], c[3010], d[3010], sum=0.0;8scanf"%d", &T);9 while(t--)Ten { Onescanf"%d", &n); Ascanf"%LF%LF", &x, &y); - for(i =1; I <= N; i++) - { thescanf"%LF", &b[i]); - } - for(i =1; I <= N; i++) - { +scanf"%LF", &c[i]); -D[i] = b[i]-C[i]; + } A for(i =1; I <= N; i++) at { -Sum + =2* (N-i +1) *D[i]; - } -A1 = (n * x-sum + y)/(n +1); -printf"%.2lf\n", A1); - } in - return 0; to}
AC Code
Freshman race 1011 Singles Day gift of the world