重复局面

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#include<bits/stdc++.h>

using namespace std;

map<string,int>m;

int main()
{
    int n;
    cin>>n;
    
    string s[110];
    
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=8;j++)
        {
            string tep;
            cin>>tep;
            s[i]+=tep;
        }
        if(!(m.count(s[i])))
        {
            m[s[i]]=1;
            cout<<1<<'\n';
        }
        else 
        {
            m[s[i]]++;
            cout<<m[s[i]]<<'\n';
        }
    }
    return 0;
}

矩阵运算

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#include<bits/stdc++.h>

using namespace std;

#define int long long

const int N = 1e4 + 10;
const int D = 25;

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    int n, d;
    int Q[N][D], K[N][D], V[N][D], w[N], tep[N][D] = { 0 }, end[N][D] = { 0 };
    cin >> n >> d;

    //cin juzhen
    for (int i = 1; i <= n; i++)for (int j = 1; j <= d; j++)cin >> Q[i][j];
    for (int i = 1; i <= n; i++)for (int j = 1; j <= d; j++)cin >> K[i][j];
    for (int i = 1; i <= n; i++)for (int j = 1; j <= d; j++)cin >> V[i][j];

    for (int i = 1; i <= n; i++)cin >> w[i];

    //K^T*V
    //memset(tep,1,sizeof(tep));
    for (int i = 1; i <= d; i++)
        for (int j = 1; j <= d; j++)
        {
            for (int m = 1; m <= n; m++)
                tep[i][j] += K[m][i] * V[m][j];
        }

    //q*tep
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= d; j++)
        {
            for (int m = 1; m <= d; m++)
            {
                end[i][j] += Q[i][m] * tep[m][j];
            }
        }


    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= d; j++)
        {
            cout << end[i][j] * w[i] << " \n"[j == d];
        }
    }
    return 0;
}

解压缩

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

#define N 2000007

int b[N], tot;

int Data[N * 10];

void add(int o, int l) {
    if (o >= l) {
        int p = tot - o + 1;
        for (int i = 1; i <= l; ++i) Data[++tot] = Data[p++];
    }
    else {
        int pp = tot - o + 1;
        int p = pp;
        for (int i = 1; i <= l; ++i) {
            Data[++tot] = Data[p++];
            if (i % o == 0) p = pp;
        }
    }
}

void pt(int x) {
    if (x < 10) putchar(x + '0');
    else putchar(x + 'a' - 10);
}

void print(int x) { pt(x >> 4); pt(x & 15); }

int main() {
    int n; scanf("%d", &n);
    bool f = 0;
    for (int i = 1, x; i <= n; ++i) {
        char c = getchar();
        while (!isdigit(c) && !isalpha(c)) c = getchar();
        x = isdigit(c) ? (c - '0') : (c - 'a' + 10);
        c = getchar();
        x = (x << 4) + (isdigit(c) ? (c - '0') : (c - 'a' + 10));
        if (f) b[++tot] = x;
        if ((x & (1 << 7)) == 0) f = true;
        //print(x); puts("!");
    }
    n = tot; tot = 0;
    //printf("%d\n", n);
    for (int i = 1; i <= n; ++i) {
        if ((b[i] & 3) == 0) { // data
            int l = b[i] >> 2;
            if (l >= 60) {
                int tmpk = l - 59; l = 0;
                for (int j = i + tmpk; j > i; --j) l = (l << 8) + b[j];
                i += tmpk;
            }
            ++l;
            //printf("%d\n", l);
            for (; l; --l) Data[++tot] = b[++i];
        }
        else if ((b[i] & 3) == 1) {
            int l = ((b[i] >> 2) & 7) + 4;
            int o = ((b[i] >> 5) << 8);
            o += b[++i];
            //printf("%d %d\n", l, o);
            add(o, l);
        }
        else {
            int l = (b[i] >> 2) + 1;
            int o = (b[i + 2] << 8) + b[i + 1];
            i += 2;
            //printf("%d %d\n", l, o);
            add(o, l);
        }
    }
    for (int i = 1; i <= tot; ++i) {
        print(Data[i]);
        if (i % 8 == 0) puts("");
    }
    return 0;
}