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CF 264A(向内的双向队列)

C. Escape from Stones
time limit per test 2 seconds
memory limit per test 256 megabytes
input standard input
output standard output
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].
You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls. www.zzzyk.com
Input
The input consists of only one line. The only line contains the string s (1 ≤ |s| ≤ 106). Each character in s will be either "l" or "r".
Output
Output n lines — on the i-th line you should print the i-th stone's number from the left.
Sample test(s)
input
llrlr
output
3
5
4
2
1
input
rrlll
output
1
2
5
4
3
input
lrlrr
output
2
4
5
3
1
Note
In the first example, the positions of stones 1, 2, 3, 4, 5 will be , respectively. So you should print the sequence: 3, 5, 4, 2, 1.
 
不能用模拟double+除法,会爆精度啊!!(long double 也不行)
其实只要根据性质,在序列前后添加即可。
靠,人生中的处女Hack,竟然是被Hack…(受?)
 
[cpp]  
#include<cstdio>  
#include<iostream>  
#include<cstdlib>  
#include<cstring>  
#include<cmath>  
#include<functional>  
#include<algorithm>  
#include<cctype>  
using namespace std;  
#define MAXN (1000000+10)  
//pair<double,int> a[MAXN];  
char s[MAXN];  
int n,a[MAXN];  
int main()  
{  
    scanf("%s",&s);  
    n=strlen(s);  
    int l=1,r=n;  
    for (int i=0;i<n;i++)  
    {  
        if (s[i]=='l') a[r--]=i+1;  
        else a[l++]=i+1;                  
    }  
      
      
      
    for (int i=1;i<=n;i++) cout<<a[i]<<endl;  
      
      
      
}  
 
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