CodeForces Round #555 Div.3

A. Reachable Numbers

代码：

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

const int maxn = 1e9 + 7;
int N;
set<int> s;

int main() {
    scanf("%d", &N);
    while(s.find(N) == s.end()) {
        s.insert(N);
        N += 1;
        while(N % 10 == 0) N /= 10;
    }

    printf("%d
", (int) s.size());

    return 0;
}

B. Long Number

代码：

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

int N;
string s;
int a[10];

int main() {
    scanf("%d", &N);
    cin >> s;
    for(int i = 1; i <= 9; i ++) {
        int x;
        scanf("%d", &x);
        a[i] = x;
    }

    for(int i = 0; i < N; i ++) {
        if(a[s[i] - '0'] > s[i] - '0') {
            for(int j = i; j < N; j ++) {
                if(a[s[j] - '0'] >= s[j] - '0')
                    s[j] = a[s[j] - '0'] + '0';
                else break;
            }
            break;
        }
    }

    cout << s << endl;

    return 0;
}

C1. Increasing Subsequence (easy version)

代码：

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

const int maxn = 2e5 + 10;
int N;
int a[maxn];

int main() {
    scanf("%d", &N);
    for(int i = 0; i < N; i ++)
        scanf("%d", &a[i]);
    string ans = "";

    int l = 0, r = N - 1, pos = 0;
    while(l <= r) {
        if(a[l] < a[r]) {
            if(a[l] > pos) {
                pos = a[l];
                l ++;
                ans += 'L';
            } else if(a[r] > pos) {
                pos = a[r];
                r --;
                ans += 'R';
            } else break;
        } else {
            if(a[r] > pos) {
                pos = a[r];
                r --;
                ans += 'R';
            } else if(a[l] > pos) {
                pos = a[l];
                l ++;
                ans += 'L';
            } else break;
        }
    }

    printf("%d
", ans.length());
    cout << ans << endl;

    return 0;
}

C2. Increasing Subsequence (hard version) 

代码（在两边一样的情况下只能选左或右跑了）：

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

const int maxn = 2e5 + 10;
int N;
int pos = 0;
int a[maxn];
string ans1 = "", ans2 = "", ans = "";

string Left(int st, int en, string s) {
    string t = s;
    int l = st, r = en;
    while(l <= r) {
        if(a[l] > pos) {
            pos = a[l];
            l ++;
            t += 'L';
        } else break;
    }
    return t;
}

string Right(int st, int en, string s) {
    string t = s;
    int l = st, r = en;
    while(l <= r) {
        if(a[r] > pos) {
            pos = a[r];
            r --;
            t += 'R';
        } else break;
    }
    return t;
}

int main() {
    scanf("%d", &N);
    for(int i = 0; i < N; i ++)
        scanf("%d", &a[i]);

    int l = 0, r = N - 1;
    int tp = 0;
    while(l <= r) {
        if(a[l] < a[r]) {
            if(a[l] > pos) {
                pos = a[l];
                l ++;
                ans += 'L';
            } else if(a[r] > pos) {
                pos = a[r];
                r --;
                ans += 'R';
            } else break;
        } else if(a[l] > a[r]) {
            if(a[r] > pos) {
                pos = a[r];
                r --;
                ans += 'R';
            } else if(a[l] > pos) {
                pos = a[l];
                l ++;
                ans += 'L';
            } else break;
        } else {
            int poo = pos;
            ans1 = ans, ans2 = ans;
            ans1 = Left(l, r, ans);
            pos = poo;
            ans2 = Right(l, r, ans);
            if(ans1.length() > ans2.length())
                //cout << ans1 << endl;
                ans = ans1;
            else //cout << ans2 << endl;
                ans = ans2;
            break;
        }
    }

    printf("%d
", ans.length());
    cout <<ans <<endl;

    return 0;
}





/*

15
37504 79054 80071 95721 135743 164345 189260 190810 191657 196168 200000 200000 190810 190018 185437

*/

E. Minimum Array

代码：

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

const int maxn = 2e5 + 10;
int N;
int a[maxn], c[maxn];

int main() {
    multiset<int> s;
    scanf("%d", &N);
    for(int i = 1; i <= N; i ++)
        scanf("%d", &a[i]);
    for(int i = 1; i <= N; i ++) {
        int x;
        scanf("%d", &x);
        s.insert(x);
    }
    for(int i = 1; i <= N; i ++) {
        int now = N - a[i] % N;
        multiset <int>::iterator it = s.lower_bound(now);
        if(it == s.end()) it = s.begin();
        c[i] = (a[i] + (*it)) % N;
        s.erase(it);
    }

    for(int i = 1; i <= N; i ++)
        printf("%d%s", c[i], i != N ? " " : "
");

    return 0;
}

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