B - Symmetric Matrix

思路：将矩阵转换成图的形式，然后推公式。

#include
      
       #define LL long long#define fi first#define se second#define mk make_pair#define pii pair
       
        #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\n", dp[n]);    }    return 0;}  /*3 1000000000100000 1000000000 507109376*/
       
      

 

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
      
       #define LL long long#define fi first#define se second#define mk make_pair#define pii pair
       
        #define y1 skldjfskldjg#define y2 skldfjsklejgusing 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\n", ans);    }    return 0;}/*3 2 21 2 14 2 21 2 1 2*/
       
      

 

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
      
       #define LL long longusing 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] = '\0';    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("\n");        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\n", ret, sa[i], last);            ans += ret;        }        printf("%lld\n", ans / 6);    }    return 0;}/*4abab4abaa*/