题目链接:点击打开链接
题意:
给定n个加油站,一辆车由A点跑到B点,每个100m有一个加油站,每开100m需要10升油。
在每个车站会检查一下油量,若车子若开不到下一个加油站则加x升油。
开始有x升油
下面给出加油的记录。
问下一次加油在哪一站。若答案唯一输出具体哪站。
油箱容量无限
思路:
水模拟。。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <iostream>
using namespace std;
#define gg 10.0
#define N 1005
#define eps (1e-6)
#define ll int
double p[N];
ll n;
void input(){
for(ll i = 1; i <= n; i++)cin>>p[i];
}
double ok(double x){
double now = x;
double siz = 0;
for(ll i = 1; i <= n; i++)
{
siz += floor(now/gg);
now -= floor(now/gg)*gg;
if(siz != p[i]){
double cha = (p[i] - siz)*10.0 - now;
now = 0;
x += cha / i;
}
siz = p[i];
now += x;
}
return now;
}
double ok2(double x){
double now = x;
double siz = 0;
for(ll i = 1; i <= n; i++)
{
siz += floor(now/gg);
now -= floor(now/gg)*gg;
if(siz != p[i]){
double cha = (siz - p[i])*10.0 - (10.0-now-eps);
now = 10.0 - eps;
x -= cha / i;
}
siz = p[i];
now += x;
}
return now;
}
int main(){
ll i, j;
while(cin>>n){
input();
double l = ok(p[1]*10.0) + eps;
double r = ok2(p[1]*10.0 + 10.0);
l = floor(l/10.0);
r = floor(r/10.0);
if(l == r)
{
puts("unique");
printf("%.0lf\n", p[n] + l);
}
else puts("not unique");
}
return 0;
}