cf605 div3

A题。。年轻人老想不一样的姿势。。。折腾半天直接枚举不香吗？呸，真香
B题。。。没仔细看题 WA到自闭

C题。。题目都告诉公式 n*(n+1)/2 了，折腾半天发现不是 2^n-1不是组合数-{n，0} 然后推出题目给的公式可还行，对自己服气哦可真是棒呢

D题。。。有思路但是没写出来，时间不够。。。。

A题，直接枚举吧，已有几种不一样的姿势提供

B题

//Recently you have bought a snow walking robot and brought it home
//so the rebot should come back to the home(0, 0)
//and notice you can't visit one point twice(except 0,0)
// ans += string(x, 'c'); //string copy x chars 'c'

C题，模拟跑一遍

#include <bits/stdc++.h>//cf605 div3
#define ll long long
#define _rep(i,a,b) for(int i = (a); i <= (b) ;i++)
#define _for(i,a,b) for(int i = (a); i < (b) ;i++)
using namespace std;

void taskA(){
    int q;
    cin >> q;
    while(q--){
        int a[4]; 
        cin >> a[1] >> a[2] >> a[3];
        sort(a+1, a+4);
        ll ans = a[2]-a[1]+a[3]-a[1]+a[3]-a[2];
        if(a[3]-a[1]<=2) ans = 0;
        for(int i = 4; i > 0; i--)
            if(ans-i > 0) {ans -= i; break;}
        cout << ans << endl;
    }return;
}

void taskA1(){
    int q,a,b,c;
    cin >> q;
    while(q--){
        cin >> a >>b >> c;    
        ll ans= 1e18;
        _rep(i,-1,1) _rep(j,-1,1) _rep(k,-1,1){
            int x = a+i;
            int y = b+j;
            int z = c+k;
            ans = min(ans, (ll)(abs(x-y)+abs(y-z)+abs(x-z)));
        }   
        cout << ans << endl;
    }return;
}
void taskA2(){
    int t; cin >> t;
    while(t--){
        int a[4];
        cin >> a[1] >> a[2] >> a[3];
        sort(a+1, a+4);
        int mx = max(0, a[3]-a[1]-2);
        cout << mx*2 << endl;
    }return;
}

View Code

void taskB(){
    int t; cin >> t;
    while(t--){
        string s;
        cin >> s;
        int a[27] = {}; 
        for(char ch : s) a[ch-'A']++;
        int cl = min(a['L'-'A'], a['R'-'A']);
        int cu = min(a['U'-'A'], a['D'-'A']);
        if(!cl || !cu) cl = min(cl, 1), cu = min(cu, 1);
        string ans = "";
    //cout << "cl = " << cl << "  cu = " << cu << endl;
        ans += string(cu, 'U');//_for(i,0,cu) ans += 'U';
        ans += string(cl, 'L');//_for(i,0,cl) ans += 'L';
        ans += string(cu, 'D');//_for(i,0,cu) ans += 'D';
        ans += string(cl, 'R');//_for(i,0,cl) ans += 'R';
        
        if(ans.size() == 0){ cout << "0\n\n"; continue;}
        cout << ans.size() << endl << ans << endl;
    }return;
}

void taskC(){
    int n,k; cin >> n >> k;
    string s; cin >> s;
    vector<int> used(27, 0);
    for(int i = 0; i < k; i++) {
        char ch; cin >> ch; cin.get();
        used[ch-'a'] = 1;
        //cout << "ch = " << ch << endl;
    }
    ll ans = 0, cnt = 0;
    for(int i = 0; i < n; i++) {
        if(!used[s[i]-'a']) {
            if(cnt > 0) ans += (cnt*(cnt+1)/2);
            cnt = 0;
        }
        else cnt++;
    }
    if(cnt > 0) ans += (cnt*(cnt+1)/2);
    cout << ans << endl;
    return;
}

void taskC1(){
    int n,k;
    cin >> n >>k;
    string s; cin >> s;
    vector<int> v(27,0);
    _for(i,0,k) { char ch;; cin >> ch; cin.get();v[ch-'a'] = 1;}

    ll ans = 0;
    _for(i,0,n){
        int pos = i;
        while(v[s[i]-'a']) i++;
        ans += 1LL*(i-pos)*(i-pos+1)/2;
    }cout << ans << endl;
    return;
}

View Code

int main(){
    //taskA();
    taskB();
    //taskC();
    //taskC1();
    //taskD();
    return 0;
}

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