• [Offer收割]编程练习赛39


    公平分队

    #pragma comment(linker, "/STACK:102400000,102400000")
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<vector>
    #include<algorithm>
    #include<iostream>
    #include<map>
    #include<queue>
    #include<stack>
    #include<string>
    #include<functional>
    //#include<bits/stdc++.h>
    using namespace std;
    typedef long long lint;
    typedef vector<int> VI;
    typedef pair<int, int> PII;
    
    
    
    int cmp(const void * x, const void * y) {
    #define datatype int
        datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));
        //x < y
        return dx < dy ? -1 : 1;
    #undef datatype
    }
    
    int a[200005];
    
    int main() {
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
    #endif
        std::ios::sync_with_stdio(0), cin.tie(0);
        int n;
        cin >> n;
        for (int i = 0; i < 2 * n; i++) cin >> a[i];
        sort(a, a + 2 * n);
        lint ans = 0;
        for (int i = 1; i < n; i++) ans += a[i];
        ans += a[2 * n - 1];
        cout << ans << endl;
        return 0;
    }
    View Code

    XY游戏

    #pragma comment(linker, "/STACK:102400000,102400000")
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<vector>
    #include<algorithm>
    #include<iostream>
    #include<map>
    #include<queue>
    #include<stack>
    #include<string>
    #include<functional>
    //#include<bits/stdc++.h>
    using namespace std;
    typedef long long lint;
    typedef vector<int> VI;
    typedef pair<int, int> PII;
    
    
    
    int cmp(const void * x, const void * y) {
    #define datatype int
        datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));
        //x < y
        return dx < dy ? -1 : 1;
    #undef datatype
    }
    
    map<int, int> mp;
    struct state {
        int a[4][4];
        int step;
    };
    queue<state> q;
    int hash_state(state st) {
        int rtn = 0;
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 4; j++)
                rtn = rtn * 3 + st.a[i][j];
        }
        return rtn;
    }
    const int dx[4] = {-1, 0, 1, 0};
    const int dy[4] = {0, 1, 0, -1};
    int main() {
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
    #endif
        std::ios::sync_with_stdio(0), cin.tie(0);
        state initial, st, stt;
        char ch;
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 4; j++) {
                cin >> ch;
                while (ch != 'O' && ch != 'X' && ch != 'Y') cin >> ch;
                if (ch == 'O') initial.a[i][j] = 0;
                else if (ch == 'X') initial.a[i][j] = 1;
                else initial.a[i][j] = 2;
            }
        }
        initial.step = 0;
        q.push(initial);
        mp[hash_state(initial)]++;
        while (!q.empty()) {
            st = q.front();
            q.pop();
            bool ok = false;
            for (int i = 0; i < 4; i++) {
                if (st.a[i][0] == 1 && st.a[i][1] == 1 && st.a[i][2] == 1 && st.a[i][3] == 1) ok = true;
                if (st.a[i][0] == 2 && st.a[i][1] == 2 && st.a[i][2] == 2 && st.a[i][3] == 2) ok = true;
                if (st.a[0][i] == 1 && st.a[1][i] == 1 && st.a[2][i] == 1 && st.a[3][i] == 1) ok = true;
                if (st.a[0][i] == 2 && st.a[1][i] == 2 && st.a[2][i] == 2 && st.a[3][i] == 2) ok = true;
                if (st.a[0][0] == 1 && st.a[1][1] == 1 && st.a[2][2] == 1 && st.a[3][3] == 1) ok = true;
                if (st.a[0][0] == 2 && st.a[1][1] == 2 && st.a[2][2] == 2 && st.a[3][3] == 2) ok = true;
                if (st.a[0][3] == 1 && st.a[1][2] == 1 && st.a[2][1] == 1 && st.a[3][0] == 1) ok = true;
                if (st.a[0][3] == 2 && st.a[1][2] == 2 && st.a[2][1] == 2 && st.a[3][0] == 2) ok = true;
            }
            if (ok) {
                cout << st.step << endl;
                return 0;
            }
            stt = st;
            stt.step = st.step + 1;
            for (int i = 0; i < 4; i++) {
                for (int j = 0; j < 4; j++) {
                    if (st.a[i][j] == 0) continue;
                    for (int k = 0; k < 4; k++) {
                        if (0 <= i + dx[k] && i + dx[k] < 4 && 0 <= j + dy[k] && j + dy[k] < 4 && st.a[i + dx[k]][j + dy[k]] == 0) {
                            stt.a[i + dx[k]][j + dy[k]] = stt.a[i][j];
                            stt.a[i][j] = 0;
                            int h = hash_state(stt);
                            if (mp.find(h) == mp.end()) {
                                mp[h]++;
                                q.push(stt);
                            }
                            stt.a[i][j] = stt.a[i + dx[k]][j + dy[k]];
                            stt.a[i + dx[k]][j + dy[k]] = 0;
                        }
                    }
                }
            }
        }
        cout << -1 << endl;
        return 0;
    }
    View Code

    第K小最简真分数

    #pragma comment(linker, "/STACK:102400000,102400000")
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<vector>
    #include<algorithm>
    #include<iostream>
    #include<map>
    #include<queue>
    #include<stack>
    #include<string>
    #include<functional>
    //#include<bits/stdc++.h>
    using namespace std;
    typedef long long lint;
    typedef vector<int> VI;
    typedef pair<int, int> PII;
    
    
    
    int cmp(const void * x, const void * y) {
    #define datatype int
        datatype dx = *((datatype *)(x)), dy = *((datatype *)(y));
        //x < y
        return dx < dy ? -1 : 1;
    #undef datatype
    }
    lint p[100];
    int m = 0;
    lint calc(lint x) {
        lint rtn = x;
        for (int i = 1; i < (1 << m); i++) {
            int cnt = 0, mlt = 1;
            for (int j = 0; j < m; j++) {
                if (i & (1 << j)) {
                    cnt++;
                    mlt *= p[j];
                }
            }
            if (cnt & 1) rtn -= x / mlt;
            else rtn += x / mlt;
        }
        return rtn;
    }
    int main() {
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
    #endif
        std::ios::sync_with_stdio(0), cin.tie(0);
        lint n, k;
        cin >> n >> k;
        lint l = 1, r = n;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                p[m++] = i;
                while (n % i == 0) n /= i;
            }
        }
        if (n != 1) p[m++] = n;
        while (l < r) {
            lint mid = (l + r) >> 1;
            lint fk = calc(mid);
            if (fk < k) l = mid + 1;
            else r = mid;
        }
        cout << r << endl;
        return 0;
    }
    View Code

    前缀后缀查询

    #include <bits/stdc++.h>
    
    using namespace std;
    #define pii pair<int,int>
    #define vi vector<int>
    #define pb push_back
    #define mp make_pair
    #define x first
    #define y second
    #define SZ(x) x.size()
    #define all(x) x.begin(),x.end()
    #define rep(i,a,b) for(int i=a;i<b;i++)
    #define per(i,a,b) for(int i=b-1;i>=a;i--)
    #define DBG(x) cerr<<(#x)<<"="<<x<<"
    ";
    #define inf 1000000007
    #define mod 1000000007
    #define ll long long 
    #define N 100010
    
    template<class T,class U> void Max(T &a,U b){if(a<b)a=b;}
    template<class T,class U> void Min(T &a,U b){if(a>b)a=b;}
    template<class T,class U> void add(T &a,U b){a+=b;if(a>=mod)a-=mod;}
    
    template<class T,class U> int min(T a,U b){return a>=b?b:a;}
    
    int a[12],b[12],sz;
    char s[12];
    int ch[1000007][26];
    void add(char s[],int a[]){
        int n=strlen(s),p=0;
        rep(i,0,n){
            int j=s[i]-'a';
            if(!ch[p][j])ch[p][j]=++sz;
            p=ch[p][j];
            a[i+1]=p;
        }
    }
    int find(string s){
        int p=0;
        rep(i,0,SZ(s)){
            int j=s[i]-'a';
            if(!ch[p][j])return -1;
            p=ch[p][j];
        }
        return p;
    }
    int main(){
        int i,j,k,ca=0,T,n,m,K;
        scanf("%d%d",&n,&K);
        unordered_map<ll,int>g;
        int w;sz=0;
        rep(i,0,n){
            scanf("%s%d",s,&w);
            m=strlen(s);
            add(s,a);
            reverse(s,s+m);
            add(s,b);
            rep(j,0,m+1){
                rep(k,0,m+1){
                    ll x=a[j]*1000007LL+b[k];
                    if(!g.count(x))g[x]=w;
                    else if(w>g[x])g[x]=w;
                }
            }
        }
        string s,t;
        while(K--){
            cin>>s>>t;
            reverse(all(t));
            int ans=-1;
            int x=find(s),y=find(t);
            if(x!=-1&&y!=-1){
                ll w=x*1000007LL+y;
                if(g.count(w))ans=g[w];
            }
            printf("%d
    ",ans);
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/dramstadt/p/8021933.html
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