牛客网暑期多校1

B - Symmetric Matrix

思路：将矩阵转换成图的形式，然后推公式。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg
 
using namespace std;
 
const int N = 1e5 + 7;
const int M = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
 
LL dp[N], sum[N];
int mod, n;
LL fastPow(LL a, LL b) {
    LL ans = 1;
    while(b) {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod; b >>= 1;
    }
    return ans;
}
 
void init() {
 
    dp[1] = 0;
    dp[2] = 1;
    dp[0] = 1;
    for(int i = 3; i <= n; i++) {
        sum[i] = (i - 1) * sum[i - 1] % mod + (1ll * (i - 1) * (i - 2) / 2) % mod * dp[i - 3] % mod;
        if(sum[i] >= mod) sum[i] -= mod;
 
        dp[i] = (i - 1) * dp[i - 2] % mod + sum[i];
        if(dp[i] >= mod) dp[i] -= mod;
 
    }
}
 
int main() {
    while(scanf("%d%d", &n, &mod) != EOF) {
        init();
        printf("%lld
", dp[n]);
    }
    return 0;
}
 
 
/*
3 1000000000
100000 1000000000
 
507109376
*/

View Code

E - Removal

题目大意：给你 n 个数，每个数在1 - k 之间，现在让你删除 m 个数，问你最后有多少种序列。

思路：存在重复的问题，对于同一个序列我们只记录最靠前的一个，我们用nx[ i ][ j ]表示 i 后边第一个j 的位置。

可以得到dp方程：每一个dp[ i ][ j ] 可以转移到 dp[ nx[ i ][ k ] ][ j + nx[ i ][ k ] - i - 1]

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N = 1e5 + 7;
const int M = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +7;

int n, m, up, a[N], mp[11], nx[N][11], dp[N][11];

void init() {
    memset(mp, 0, sizeof(mp));
    for(int i = 0; i <= n; i++)
        for(int j = 0; j <= 10; j++)
            dp[i][j] = nx[i][j] = 0;
}

void add(int &a, int b) {
    a += b; if(a >= mod) a -= mod;
}
int main() {

    while(scanf("%d%d%d", &n, &m, &up) != EOF) {

        init();

        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }

        for(int i = n; i >= 0; i--) {
            for(int j = 1; j <= up; j++) {
                nx[i][j] = mp[j];
            }
            mp[a[i]] = i;
        }

        dp[0][0] = 1;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j <= m; j++) {

                if(!dp[i][j]) continue;

                for(int k = 1; k <= up; k++) {
                    int pos = nx[i][k];

                    if(pos && j + pos - i - 1 <= m) {
                        add(dp[pos][j + pos - i - 1], dp[i][j]);
                    }
                }
            }
        }

        int ans = 0;
        for(int i = 1; i <= n; i++) {
            int j = n - i;
            if(j <= m) add(ans, dp[i][m - j]);
        }

        printf("%d
", ans);
    }
    return 0;
}


/*
3 2 2
1 2 1
4 2 2
1 2 1 2
*/

View Code

I - Substring

题目大意：给你一个由a, b, c三种字符组成的字符串，两个子串要是形式一样就看成一种子串，比如 aba, aca, bab, bcb, cac, cbc是一种，

aaa, bbb, ccc 是一种。

思路：将 3! 种形式的字符串接在一起形成一个新的串，然后统计这个串有多少不同的子串，设单一字符子串的数量为cnt1，

非单一字符子串的数量为 cnt2，所以答案为(cnt2 + 2 * cnt1) / 6

统计有多少不同子串可以用后缀数组求。

#include<bits/stdc++.h>
#define LL long long

using namespace std;

const int N = 3e5 + 7;

char s[N], str[N];
int sa[N], t[N], t2[N], c[N], rk[N], height[N], id[N], b[N], d[N], n, tot;

void buildSa(char *s, int n, int m) {
    int i, j = 0, k = 0, *x = t, *y = t2;
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) c[x[i] = s[i]]++;
    for(i = 1; i < m; i++) c[i] += c[i - 1];
    for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
    for(int k = 1; k <= n; k <<= 1) {
        int p = 0;
        for(i = n - k; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = 0; i < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[y[i]]]++;
        for(i = 1; i < m; i++) c[i] += c[i - 1];
        for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for(int i = 1; i < n; i++) {
            if(y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k])
                x[sa[i]] = p - 1;
            else x[sa[i]] = p++;
        }
        if(p >= n) break;
        m = p;
     }

     for(i = 1; i < n; i++) rk[sa[i]] = i;
     for(i = 0; i < n - 1; i++) {
        if(k) k--;
        j = sa[rk[i] - 1];
        while(s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
     }
}

void init() {
    b[n] = 0; tot = 0;
    for(int i = n - 1; i >= 0; i--) {
        if(s[i] == s[i + 1]) b[i] = b[i + 1] + 1;
        else b[i] = 1;
        for(int j = 1; j <= 5; j++) b[j * (n + 1) + i] = b[i];
    }

    id[0] = 0; id[1] = 1; id[2] = 2;
    char ch = 'A';
    do {
        for(int i = 0; i < n; i++) {
            str[tot++] = id[s[i] - 'a'] + 'a';
        }

        str[tot++] = ch++;
    } while(next_permutation(id, id + 3));

    str[tot] = '';

    int pos = -1;
    for(int i = tot - 1; i >= 0; i--) {
        if(b[i] == 0) pos = i;
        d[i] = pos;
    }
}
int main() {
    while(scanf("%d", &n) != EOF) {
        scanf("%s", s);
        init();

//        puts("");
//        puts(str);
//        puts("
");
        buildSa(str, tot + 1, 180);

        LL ans = 0;
        for(int i = 1; i <= tot; i++) {
//            puts(str + sa[i]);
            int lcp = height[i];
            int last = d[sa[i]];
            int cnt = b[sa[i]];
            LL ret = 0;
            if(sa[i] == last || lcp >= last - sa[i]) continue;

            if(cnt > lcp) {
                ret += 2 * (cnt - lcp);
                ret += last - sa[i] - cnt;
            } else {
                ret += last - sa[i] - lcp;
            }

//            printf("%lld   %d   %d
", ret, sa[i], last);
            ans += ret;
        }

        printf("%lld
", ans / 6);
    }
    return 0;
}
/*
4
abab

4
abaa
*/

View Code