2017ccpc女生赛

6023 模拟

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int T;
int cnt[2010];
int vis[2010];

int main()
{
    int n, m;
    cin >> T;

    while(T--)
    {
        memset(vis, 0, sizeof(vis));
        memset(cnt, 0, sizeof(cnt));
        int a, b, c;
        char ch, s[6];
        cin >> n >> m;
        int times = 0;
        int sum = 0;
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d%c%d%s", &a, &b, &ch, &c, s);
            if(vis[a])  continue;

            if(s[0] == 'A')
            {
                vis[a] = 1;
                int tmp = b * 60 + c;
                times++;
                sum += cnt[a] * 20 + tmp;
            }
            else
            {
                cnt[a]++;
            }
        }
        cout << times << " " << sum << endl;
    }
    return 0;
}

6025 前缀和思想

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>
#include<map>
using namespace std;
const int maxn = 2e5 + 10;
int T;
int a[maxn];
//int cnt[maxn];
int forw[maxn], backw[maxn];


int gcd(int a, int b)
{
    if(b == 0)
        return a;
    else
        return gcd(b, a % b);
}


int main()
{
    cin >> T;
    int n;
    int t;
    //cout << gcd(1, 1) << endl;

    for(int kase = 1; kase <= T; kase++)
    {
        scanf("%d", &n);
        //cout << "ford" << endl;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            if(i == 1)
                forw[i] = a[i];
            else
            {
                forw[i] = gcd(forw[i - 1], a[i]);
            }
            //cout << forw[i] << endl;
        }
        //cout << "backw" << endl;
        for(int i = n; i >= 1; i--)
        {
            if(i == n)
                backw[i] = a[i];
            else
            {
                backw[i] = gcd(backw[i + 1], a[i]);
            }
            //cout << backw[i] << endl;
        }

        int mmax = max(backw[2], forw[n - 1]);
        for(int i = 2; i <= n - 1; i++)
        {
            mmax = max(mmax, gcd(forw[i - 1], backw[i + 1]));
        }
        cout << mmax << endl;
    }
    return 0;
}

6027 基础取模

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;


ll qpow(ll a, ll b)
{
    ll ans = 1;
    while(b)
    {
        if(b & 1)
            ans = ans * a % MOD;
        b >>= 1;
        a = a * a % MOD;
    }
    return ans;
}

ll cal(ll n, int k)
{
    ll sum = 0;
    for(ll i = 1; i <= n; i++)
    {
        sum = (sum + qpow(i, k)) % MOD;
    }
    return sum;
}


int main()
{
    int T;
    cin >> T;
    ll a;
    int b;
    while(T--)
    {
        cin >> a >> b;
        cout << cal(a, b) << endl;
    }
   return 0;
}

6030 递推+矩阵快速幂（构造矩阵）

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
struct Mat
{
    ll m[5][5];

};


Mat mul(Mat a, Mat b)
{
    Mat c;
    memset(c.m, 0, sizeof(c.m));
    for(int i = 1; i <= 4; i++)
    {
        for(int j = 1; j <= 4; j++)
        {
            for(int k = 1; k <= 4; k++)
            {
                c.m[i][j] = (c.m[i][j] + (a.m[i][k] * b.m[k][j])) % MOD;
            }
        }
    }
    return c;
}


void calc(Mat ori, ll n, Mat &res)      //take care ll of n
{
    //Mat res;
    memset(res.m, 0, sizeof(res.m));
    res.m[1][1] = res.m[2][2] = res.m[3][3] = res.m[4][4] = 1;

    while(n)
    {
        if(n & 1)
        {
            res = mul(res, ori);
        }
        ori = mul(ori, ori);
        n >>= 1;
    }
}

int main()
{
    int T;
    cin >> T;
    ll n;
    while(T--)
    {
        scanf("%lld", &n);
        if(n == 1)
            printf("1\n");
        else if(n == 2)
            printf("3\n");
        else if(n == 3)
            printf("4\n");
        else if(n == 4)
            printf("6\n");
        else
        {
            Mat ans;
            Mat mid;
            memset(mid.m, 0, sizeof(mid.m));
            mid.m[1][1] = mid.m[1][3] = mid.m[2][1] = mid.m[3][2] = mid.m[4][3] = 1;
            calc(mid, n - 4, ans);
            ll sum = (ans.m[1][1] * 6 + ans.m[1][2] * 4 + ans.m[1][3] * 3 + ans.m[1][4] * 1) % MOD;
            printf("%lld\n", sum);
        }
    }
    return 0;
}