ExaWizards 2019

C - Snuke the Wizard

发现符文的相对位置不变，直接二分某个位置是否到达最左或最右来计算

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define eps 1e-10

#define MAXN 200005

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

    res = 0;T f = 1;char c = getchar();

    while(c < '0' || c > '9') {

        if(c == '-') f = -1;

        c = getchar();

    }

    while(c >= '0' && c <= '9') {

        res = res * 10 +c - '0';

        c = getchar();

    }

    res *= f;

}

template<class T>

void out(T x) {

    if(x < 0) {x = -x;putchar('-');}

    if(x >= 10) {

        out(x / 10);

    }

    putchar('0' + x % 10);

}

int N,Q;

char s[MAXN];

pii op[MAXN];

int getplace(int p) {

    for(int i = 1 ; i <= Q ; ++i) {

        if(s[p] == 'A' + op[i].fi) {

            p += op[i].se;

        }

    }

    return p;

}

int findL() {

    int L = 0,R = N;

    while(L < R) {

        int mid = (L + R + 1) >> 1;

        if(getplace(mid) == 0) L = mid;

        else R = mid - 1;

    }

    return L;

}

int findR() {

    int L = 1,R = N + 1;

    while(L < R) {

        int mid = (L + R) >> 1;

        if(getplace(mid) == N + 1) R = mid;

        else L = mid + 1;

    }

    return L;

}

void Solve() {

    read(N);read(Q);

    scanf("%s",s + 1);

    char t[5],d[5];

    for(int i = 1 ; i <= Q ; ++i) {

        scanf("%s%s",t + 1,d + 1);

        int a,b;

        if(d[1] == 'L') b = -1;

        else b = 1;

        a = t[1] - 'A';

        op[i] = mp(a,b);

    }

    int L = findL(),R = findR();

    out(R - L - 1);enter;

}

int main() {

#ifdef ivorysi

    freopen("f1.in","r",stdin);

#endif

    Solve();

    return 0;

}

D - Modulo Operations

如果按照我们对值产生贡献的序列一定是一个递减的序列

我们只要按照值从大到小排序后记录\(dp[i][j]\)为到第i个数取模的数是j的概率是多少

转移的时候如果对当前值进行取模，则乘上一个\(\frac{1}{N - i + 1}\)

否则取乘一个\(\frac{N- i}{N - i + 1}\)

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define eps 1e-10

#define MAXN 200005

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

    res = 0;T f = 1;char c = getchar();

    while(c < '0' || c > '9') {

        if(c == '-') f = -1;

        c = getchar();

    }

    while(c >= '0' && c <= '9') {

        res = res * 10 +c - '0';

        c = getchar();

    }

    res *= f;

}

template<class T>

void out(T x) {

    if(x < 0) {x = -x;putchar('-');}

    if(x >= 10) {

        out(x / 10);

    }

    putchar('0' + x % 10);

}

const int MOD = 1000000007;

int N,X,a[205],fac[205],inv[205];

int dp[205][100005];

int inc(int a,int b) {

    return a + b >= MOD ? a + b - MOD : a + b;

}

int mul(int a,int b) {

    return 1LL * a * b % MOD;

}

void update(int &x,int y) {

    x = inc(x,y);

}

void Solve() {

    read(N);read(X);

    for(int i = 1 ; i <= N ; ++i) read(a[i]);

    sort(a + 1,a + N + 1,[](int a,int b){return a > b;});

    fac[0] = 1;

    for(int i = 1 ; i <= N ; ++i) fac[i] = mul(fac[i - 1],i);

    inv[1] = 1;

    for(int i = 2 ; i <= N ; ++i) inv[i] = mul(inv[MOD % i],MOD - MOD / i);

    dp[0][X] = 1;

    for(int i = 1 ; i <= N ; ++i) {

        for(int j = 0 ; j <= X ; ++j) {

            if(j < a[i]) update(dp[i][j],dp[i - 1][j]);

            else {

                update(dp[i][j % a[i]],mul(dp[i - 1][j],inv[N - i + 1]));

                update(dp[i][j],mul(dp[i - 1][j],mul(N - i,inv[N - i + 1])));

            }

        }

    }

    int ans = 0;

    for(int j = 0 ; j <= X ; ++j) {

        update(ans,mul(dp[N][j],j));

    }

    ans = mul(ans,fac[N]);

    out(ans);enter;

}

int main() {

#ifdef ivorysi

    freopen("f1.in","r",stdin);

#endif

    Solve();

    return 0;

}

E - Black or White

我们把点画在坐标轴上,每个i相当于画一条斜线，那么如果这个斜线上的点都有让B减少的一步，那么答案就是\(\frac{1}{2}\)

否则处理出剩了多少个白点，和剩了多少个黑点的情况，每条斜线最多涉及两个这样的点，把他们的贡献加上或减掉即可

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define eps 1e-10

#define MAXN 200005

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

    res = 0;T f = 1;char c = getchar();

    while(c < '0' || c > '9') {

        if(c == '-') f = -1;

        c = getchar();

    }

    while(c >= '0' && c <= '9') {

        res = res * 10 +c - '0';

        c = getchar();

    }

    res *= f;

}

template<class T>

void out(T x) {

    if(x < 0) {x = -x;putchar('-');}

    if(x >= 10) {

        out(x / 10);

    }

    putchar('0' + x % 10);

}

const int MOD = 1000000007;

int B,W;

int fac[MAXN],invfac[MAXN],pw[MAXN];

int r[MAXN],c[MAXN];

int inc(int a,int b) {

    return a + b >= MOD ? a + b - MOD : a + b;

}

int mul(int a,int b) {

    return 1LL * a * b % MOD;

}

int C(int n,int m) {

    if(n < m) return 0;

    return mul(fac[n],mul(invfac[m],invfac[n - m]));

}

int fpow(int x,int c) {

    int res = 1,t = x;

    while(c) {

        if(c & 1) res = mul(res,t);

        t = mul(t,t);

        c >>= 1;

    }

    return res;

}

void Solve() {

    read(B);read(W);

    fac[0] = 1;

    for(int i = 1 ; i <= B + W ; ++i) fac[i] = mul(fac[i - 1],i);

    invfac[B + W] = fpow(fac[B + W],MOD - 2);

    for(int i = B + W - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);

    pw[0] = 1;pw[1] = invfac[2];

    for(int i = 2 ; i <= B + W ; ++i) pw[i] = mul(pw[i - 1],pw[1]);

    r[0] = pw[W];

    for(int i = 1; i <= B ; ++i) {

        r[i] = inc(r[i - 1],mul(pw[1],mul(pw[i + W - 1],C(i + W - 1,i))));

    }

    c[0] = pw[B];

    for(int i = 1 ; i <= W ; ++i) {

        c[i] = inc(c[i - 1],mul(pw[1],mul(pw[B - 1 + i],C(B - 1 + i,i))));

    }

    for(int i = 0 ; i < B + W ; ++i) {

        int res = (MOD + 1) / 2;

        if(i >= B) {

            res = inc(res,MOD - mul((MOD + 1) / 2,c[i - B]));

        }

        if(i >= W) {

            res = inc(res,mul((MOD + 1) / 2,r[i - W]));

        }

        out(res);enter;

    }

}

int main() {

#ifdef ivorysi

    freopen("f1.in","r",stdin);

#endif

    Solve();

    return 0;

}