看完题目后，得到公式a[i]=2*a[i-1]-a[i-2]+2*c[i-1]; YY发现a[n+1]随a[1]增大而增大，那么就对a[1]二分吧，精度1e-4,1A。
Simple calculations
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5989 | Accepted: 3040 |
Description
There is a sequence of n+2 elements a0, a1, ..., an+1 (n <= 3000, -1000 <= ai <=1000). It is known that
ai = (ai-1 + ai+1)/2 - ci
for each i=1, 2, ..., n.
You are given a0, an+1, c1, ... , cn. Write a program which calculates a1.
ai = (ai-1 + ai+1)/2 - ci
for each i=1, 2, ..., n.
You are given a0, an+1, c1, ... , cn. Write a program which calculates a1.
Input
The first line of an input contains an integer n. The next two lines consist of numbers a0 and an+1 each 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.
two digits after decimal point), one number per line.
Output
The output file should contain a1 in the same format as a0 and an+1.
Sample Input
1 50.50 25.50 10.15
Sample Output
27.85
Source
#include <stdio.h>
int n;
const int S=3050;
double a[S],c[S],anp1;
double fun(){
int i;
for (i=2;i<=n+1;i++)
a[i]=2*a[i-1]-a[i-2]+2*c[i-1];
return a[n+1];
}
int main(){
int i;
double l,r,mid,tmp;
scanf("%d",&n);
scanf("%lf %lf",&a[0],&anp1);
for (i=1;i<=n;i++) scanf("%lf",&c[i]);
l=-1005;r=1005;
// a[1]=27.85;fun();printf("%.2lf ||||| \n",a[2]);
while (r-l>1e-4){
mid=(l+r)/2.0;
a[1]=mid;
tmp=fun();
if (a[n+1]>anp1) {r=mid;}else if ( a[n+1]<anp1 ) {l=mid;} else break;
}
printf("%.2lf\n",mid);
return 0;
}