• 1296E1


    1296E1 - String Coloring (easy version)

    因为只有两种颜色,只要可以排成两组非递减序列即可

    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <string>
    #include <set>
    #include <queue>
    #include <map>
    #include <sstream>
    #include <cstdio>
    #include <cstring>
    #include <numeric>
    #include <cmath>
    #include <iomanip>
    #include <deque>
    #include <bitset>
    //#include <unordered_set>
    //#include <unordered_map>
    #define ll              long long
    #define pii             pair<int, int>
    #define rep(i,a,b)      for(ll  i=a;i<=b;i++)
    #define dec(i,a,b)      for(ll  i=a;i>=b;i--)
    #define forn(i, n)      for(ll i = 0; i < int(n); i++)
    using namespace std;
    int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
    const long long INF = 0x7f7f7f7f7f7f7f7f;
    const int inf = 0x3f3f3f3f;
    const double pi = 3.14159265358979323846;
    const double eps = 1e-6;
    const int mod = 998244353;
    const int N = 2e5 + 5;
    //if(x<0 || x>=r || y<0 || y>=c)
    
    inline ll read()
    {
        ll x = 0; bool f = true; char c = getchar();
        while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
        while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
        return f ? x : -x;
    }
    ll gcd(ll m, ll n)
    {
        return n == 0 ? m : gcd(n, m % n);
    }
    ll lcm(ll m, ll n)
    {
        return m * n / gcd(m, n);
    }
    bool prime(int x) {
        if (x < 2) return false;
        for (int i = 2; i * i <= x; ++i) {
            if (x % i == 0) return false;
        }
        return true;
    }
    inline int qpow(int x, ll n) {
        int r = 1;
        while (n > 0) {
            if (n & 1) r = 1ll * r * x % mod;
            n >>= 1; x = 1ll * x * x % mod;
        }
        return r;
    }
    inline int add(int x, int y) {
        return ((x%mod)+(y%mod))%mod;
    }
    inline int sub(int x, int y) {
        x -= y;
        return x < 0 ? x += mod : x;
    }
    inline int mul(int x, int y) {
        return (1ll * (x %mod) * (y % mod))%mod;
    }
    inline int Inv(int x) {
        return qpow(x, mod - 2);
    }
    
    int main()
    {
        int n;
        string s,res;
        cin >> n >> s;
        char t1='a', t2='a';
        forn(i, n)
        {
            if (s[i] >= t1)
            {
                res += '0';
                t1 = s[i];
            }
            else if (s[i] >= t2)
            {
                res += '1';
                t2 = s[i];
            }
            else
            {
                cout << "NO" << endl;
                return 0;
            }
        }
        cout<<"YES
    " << res << endl;
        return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/dealer/p/13237847.html
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