cf Educational Codeforces Round 60 C. Magic Ship

原题：
C. Magic Ship

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

You a captain of a ship. Initially you are standing in a point (x1,y1) (obviously, all positions in the sea can be described by cartesian plane) and you want to travel to a point (x2,y2).

You know the weather forecast — the string s of length n, consisting only of letters U, D, L and R. The letter corresponds to a direction of wind. Moreover, the forecast is periodic, e.g. the first day wind blows to the side s1, the second day — s2, the n-th day — sn and (n+1)-th day — s1 again and so on.

Ship coordinates change the following way:

if wind blows the direction U, then the ship moves from (x,y) to (x,y+1);
if wind blows the direction D, then the ship moves from (x,y) to (x,y−1);
if wind blows the direction L, then the ship moves from (x,y) to (x−1,y);
if wind blows the direction R, then the ship moves from (x,y) to (x+1,y).

The ship can also either go one of the four directions or stay in place each day. If it goes then it’s exactly 1 unit of distance. Transpositions of the ship and the wind add up. If the ship stays in place, then only the direction of wind counts. For example, if wind blows the direction U and the ship moves the direction L, then from point (x,y) it will move to the point (x−1,y+1), and if it goes the direction U, then it will move to the point (x,y+2).

You task is to determine the minimal number of days required for the ship to reach the point (x2,y2).

Input
The first line contains two integers x1,y1 (0≤x1,y1≤10^9) — the initial coordinates of the ship.

The second line contains two integers x2,y2 (0≤x2,y2≤10^9) — the coordinates of the destination point.

It is guaranteed that the initial coordinates and destination point coordinates are different.

The third line contains a single integer n (1≤n≤10^5) — the length of the string s.

The fourth line contains the string s itself, consisting only of letters U, D, L and R.

Output
The only line should contain the minimal number of days required for the ship to reach the point (x2,y2).

If it’s impossible then print “-1”.

Examples
input
0 0
4 6
3
UUU
output
5
input
0 3
0 0
3
UDD
output
3
input
0 0
0 1
1
L
output

-1
Note
In the first example the ship should perform the following sequence of moves: “RRRRU”. Then its coordinates will change accordingly: (0,0) → (1,1) → (2,2) → (3,3) → (4,4) → (4,6).

In the second example the ship should perform the following sequence of moves: “DD” (the third day it should stay in place). Then its coordinates will change accordingly: (0,3) → (0,3) → (0,1) → (0,0).

In the third example the ship can never reach the point (0,1).

中文：
有一艘小船起点在(x1,y1)，可以在每个单位时刻向上下左右四个方向移动一步，或者静止不动。此外，给你一个字符串，表示当前时刻的风向，如果遍历到字符串的结尾，则重新回到字符串的开始端。风会把小船向一个方向吹动一步，如果风向与小船移动的方向相同，那么小船可以在单位时刻移动两步，其它同理，风向的字符串现在给你一个终点(x2,y2)，问你小船最少走多少步可以到终点，如果不能到终点输出-1。

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

typedef long long ll;
const int maxn=100005;

ll x[maxn],y[maxn];
ll stx,sty,sex,sey;
int n;
string s;

int main()
{
	ios::sync_with_stdio(false);
    while(cin>>stx>>sty>>sex>>sey)
    {
        cin>>n>>s;
        memset(x,0,sizeof(x));
        memset(y,0,sizeof(y));
        for(int i=1;i<=n;i++)
        {
            if(s[i-1]=='L')
            {
                x[i]=x[i-1]-1;
                y[i]=y[i-1];
            }
            if(s[i-1]=='R')
            {
                x[i]=x[i-1]+1;
                y[i]=y[i-1];
            }
            if(s[i-1]=='U')
            {
                x[i]=x[i-1];
                y[i]=y[i-1]+1;
            }
            if(s[i-1]=='D')
            {
                x[i]=x[i-1];
                y[i]=y[i-1]-1;
            }
        }

        ll L=0,R=1e18,mid,cnt,rem,tx,ty,dist;
        while(R-L>1)
        {
            mid=(L+R)>>1;

            cnt=mid/n;
            rem=mid%n;

            tx=stx+x[rem]+cnt*x[n];
            ty=sty+y[rem]+cnt*y[n];


            dist = abs(tx-sex)+abs(ty-sey);


            if(dist<=mid)
                R=mid;
            else
                L=mid;
        }
        if(R>5e17)
            R=-1;
        cout<<R<<endl;
    }
	return 0;
}

解答：

拿到题目都分析，第一想法是找下简单的规律，发现风向的字符串中向左与向右的风向可以抵消，同理上下。如果移动步数超过风向字符串的长度，可以先利用此规律计算小船可以移动到距离终点最近的点。但是鼓秋了半天没搞出来，情况太复杂-_-|||

看了题解才知道，这题需要用二分处理，做法很巧妙，每次二分需要移动的步数k，那么风吹动小船的次数也是k，先不考虑小船移动，仅看风向小船会达到哪一个位置(x3,y3)，然后判断(x3,y3)距离终点(x2,y2)能否在k步走完即可。

回想一下这道题目还是蛮简单的，没考虑到问题题目的点，怪自己太生疏…~ _ ~|||