HDU多校Round 4

Solved:3

rank:405.................................

B. Harvest of Apples

知道了S(n,m) 可以o(1)的求S(n - 1， m)，S(n + 1，m)，S(n，m - 1)，S(n，m + 1)  天秀莫队

#include <stdio.h>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + ;

int blo = ;
ll ans;
ll f[];
ll inv[];
ll anss[];

ll pow_mod(ll x, ll y)
{
    ll res = ;
    while(y)
    {
        if(y & ) res = res * x % mod;
        x = x * x % mod;
        y >>= ;
    }
    return res;
}

void prework()
{
    f[] = ; inv[] = ;
    for(int i = ; i <= ; i++) f[i] = f[i - ] * 1LL * i % mod;
    for(int i = ; i <= ; i++) inv[i] = pow_mod(f[i], mod - 2LL);
}

struct node
{
    int n, m, id;
}E[];

bool cmp(node A, node B)
{
    int a = (A.n - ) / blo;
    int b = (B.n - ) / blo;
    if(a == b) return A.m < B.m;
    else return a < b;
}

int main()
{
    ans = ;
    prework();
    int T;
    scanf("%d", &T);
    for(int i = ; i <= T; i++)
    {
        scanf("%d%d", &E[i].n, &E[i].m);
        E[i].id = i;
    }
    sort(E + , E +  + T, cmp);

    ll n = , m = ;
    for(int i = ; i <= T; i++)
    {
        while(n < E[i].n)
        {
            ans *= 2LL;
            ll tmp = f[n] * inv[m] % mod * inv[n - m] % mod;
            ans = (ans - tmp + mod) % mod;
            n++;
        }
        while(m > E[i].m)
        {
            ll tmp = f[n] * inv[m] % mod * inv[n - m] % mod;
            ans = (ans - tmp + mod) % mod;
            m--;
        }
        while(n > E[i].n)
        {
            ll tmp = f[n - ] * inv[m] % mod * inv[n -  - m] % mod;
            ans += tmp;
            ans = ans * inv[] % mod;
            n--;
        }
        while(m < E[i].m)
        {
            ll tmp = f[n] * inv[m + ] % mod * inv[n - m - ] % mod;
            ans = (ans + tmp) % mod;
            m++;
        }
        anss[E[i].id] = ans;
    }
    for(int i = ; i <= T; i++) printf("%lld\n", anss[i]);
    return ;
}

D. Nothing is Impossible

题意出锅 秒变签到题

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
typedef long long ll;

struct node
{
    int a, b;
}E[];

bool cmp(node A, node B)
{
    return A.b < B.b;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n, m;
        scanf("%d%d", &n, &m);
        for(int i = ; i <= n; i++) scanf("%d%d", &E[i].a, &E[i].b);
        sort(E + , E +  + n, cmp);

        ll tmp = ;
        int ans = n;
        for(int i = ; i <= n; i++)
        {
            if(tmp * (E[i].b + 1LL) <= m)
            {
                tmp *= (E[i].b + 1LL);
            }
            else
            {
                ans = i - ;
                break;
            }
        }
        printf("%d\n", ans);
    }
    return ;
}

J. Let Sudoku Rotate

玄学剪纸 把一个矩阵翻转的话就考虑换一种方法遍历 转四次相当于转一圈 dfs回来的时候要恢复状态

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <string.h>
using namespace std;

int ans;
int num[][];
int tmp[][];
int vis[];

int cal(char x)
{
    if(x - '' >=  && x - '' <= ) return x - '';
    else return  + x - 'A';
}

void rotate(int x, int y)
{
    int ix = x *  - ;
    int iy = y *  - ;
    for(int i1 = ix, i2 = iy; i2 <= iy + ; i2++, i1++)
        for(int j1 = iy, j2 = ix + ; j2 >= ix; j1++, j2--)
            tmp[i1][j1] = num[j2][i2];
    for(int i = ix; i <= ix + ; i++)
        for(int j = iy; j <= iy + ; j++)
            num[i][j] = tmp[i][j];
}

bool check(int x, int y)
{
    int ix = x *  - ;
    int iy = y *  - ;

    for(int i = ix; i <= ix + ; i++)
    {
        memset(vis, , sizeof(vis));
        for(int j = ; j <= y * ; j++)
        {
            if(vis[num[i][j]]) return false;
            vis[num[i][j]] = ;
        }
    }
    for(int j = iy; j <= iy + ; j++)
    {
        memset(vis, , sizeof(vis));
        for(int i = ; i <= x * ; i++)
        {
            if(vis[num[i][j]]) return false;
            vis[num[i][j]] = ;
        }
    }
    return true;
}

void dfs(int x, int y, int st)
{
    if(x == )
    {
        ans = min(ans, st);
        return;
    }

    for(int i = ; i <= ; i++)
    {
        if(i != ) rotate(x, y);
        if(check(x, y))
        {
            if(y != ) dfs(x, y + , st + i);
            else dfs(x + , , st + i);
        }
    }
    rotate(x, y);
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        ans = ;
        for(int i = ; i <= ; i++)
        {
            char s[]; scanf("%s", s + );
            for(int j = ; j <= ; j++) num[i][j] = cal(s[j]);
        }
        dfs(, , );
        printf("%d\n", ans);
    }
    return ;
}

K. Expression in Memories

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;

char s[];
int vis[];
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        memset(vis, , sizeof(vis));
        scanf("%s", s);
        int len = strlen(s);
        bool f = true;

        for(int i = ; i < len; i++)
        {
            if(s[i] == '+')
            {
                if(i ==  || i == len - ) f = false;
                if(i +  < len && s[i + ] == '+') f = false;
                if(i +  < len && s[i + ] == '*') f = false;
            }
            else if(s[i] == '*')
            {
                if(i ==  || i == len - ) f = false;
                if(i +  < len && s[i + ] == '+') f = false;
                if(i +  < len && s[i + ] == '*') f = false;
            }
            else if(s[i] - '' >  && s[i] - '' <= ) vis[i] = ;
            else if(s[i] == '')
            {
                if(i -  >=  && vis[i - ] == ) vis[i] = ;
                else
                {
                    if(i +  < len && s[i + ] - '' >=  && s[i + ] - '' <= ) f = false;
                    else if(i +  < len && s[i + ] == '?') s[i + ] = '+';
                }
            }
            else if(s[i] == '?')
            {
                if(i ==  || i +  == len) s[i] = '', vis[i] = ;
                else if(i -  >=  && vis[i - ] == )
                {
                    if(s[i - ] == '') s[i] = '+';
                    else s[i] = '', vis[i] = ;
                }
                else
                {
                    s[i] = '', vis[i] = ;
                }
            }
        }
        if(!f) puts("IMPOSSIBLE");
        else printf("%s\n", s);
    }
    return ;
}

L. Graph Theory Homework

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;

int q[];
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);

        for(int i = ; i <= n; i++) scanf("%d", &q[i]);
        int ans = sqrt(abs(q[n] - q[]));
        printf("%d\n", ans);
    }
    return ;
}