Codeforces 1117C Magic Ship (二分)

题意：

船在一个坐标，目的地在一个坐标，每天会有一个风向将船刮一个单位，船也可以移动一个单位或不动，问最少几天可以到目的地

思路：

二分天数，对于第k天

可以分解成船先被吹了k天，到达坐标（x1+sumx[k%n]+k/n*sumx[n], y1+sumy[k%n]+k/n*sumy[n]）

然后船在无风的情况下自己走k天是不是能到目的地就行了

注意二分的上限，如果能走到的情况下船可能每轮风只能移动一个有效单位，所以上限应该是1e9*1e5

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
//#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
    
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 998244353;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
//const db pi = acos(-1.0);


ll x1, x2;
ll y1, y2;
ll n;
char a[maxn];
ll sumx[maxn];
ll sumy[maxn];
bool ck(ll k){
    ll x = x1;
    ll y = y1;
    x += k/n*sumx[n] + sumx[k%n];
    y += k/n*sumy[n] + sumy[k%n];
    ll d = abs(x-x2)+abs(y-y2);
    if(d<=k)return true;
    return false;
}
int main() {
    scanf("%lld %lld", &x1,&y1);
    scanf("%lld %lld", &x2,&y2);
    scanf("%lld", &n);
    scanf("%s",a+1);
    ll l = 0, r = 1e15;
    for(int i = 1; i <= n; i++){
        sumx[i] = sumx[i-1];
        sumy[i] = sumy[i-1];
        if(a[i]=='U'){
            sumy[i]++;
        }
        if(a[i]=='D'){
            sumy[i]--;
        }
        if(a[i]=='L'){
            sumx[i]--;
        }
        if(a[i]=='R'){
            sumx[i]++;
        }
    }
    ll ans = -1;
    while(l <= r){
        ll mid = (l+r)/2;
        //printf("%lld %lld %lld\n", l, r, mid);
        if(ck(mid)){
            r = mid-1;
            ans = mid;
        }
        else{
            l = mid+1;
        }
    }
    printf("%lld", ans);
    return 0;

}
/*
10
59 -17 5 67 87 50 -71 54 27 -10

 */