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B - Escape from Stones CodeForces - 264A

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.

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.

Examples
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.

補一補昨晚的訓練題。挺長時間沒有寫部落格了,心裡滿滿的負罪感。這個題目的意思不難理解,就是一個簡單的二分。我一開始就是按照題目的要求,一點一點的分下去,樣例過了,但是提交一發就wa了。我想了想,資料量是1e6╮(╯﹏╰)╭就一個數,分1e6次也會接近零,試了試long double,還是不行。就想新的方法了。還有要注意的就是cin和cout超時,在這兒又wa了一發。
程式碼如下:

#include<bits/stdc++.h>
using namespace std;

const int maxx=1e6+10;
char a[maxx];
struct node{
	long double pos;
	int ant;
}p[maxx];

int cmp(const node &a,const node &b)
{
	return a.pos<b.pos;
}

int main()
{
	while(scanf("%s",a)!=EOF)
	{
		int len=strlen(a);
		int maxx=len;
		int minn=1;
		for(int i=0;i<len;i++)
		{
			if(a[i]=='l')
			{
				p[i].pos=maxx--;
				p[i].ant=i+1;
			}
			else
			{
				p[i].pos=minn++;
				p[i].ant=i+1;
			}
		}
		sort(p,p+len,cmp);
		for(int i=0;i<len;i++)
		printf("%d\n",p[i].ant);
		//cout<<p[i].ant<<endl;
	}
}

努力加油a啊,(o)/~