codeforces 895D

(Combinatorics)
meaning of the questions: Given two strings consisting of letters in English$s_1$,$s_2$Ensure lexicographically$s_1<s_2$, Find how many$s_3$So that$s_1<s_3<s_2$and$s_3$for$s_1$An arrangement consisting of the letters.

Ideas:
  First, statistics$s_1$What are strings of letters, each letter appears a number of times, and preserved by a num array. So that we at least know what constitutes s3 letters. Then I thought I could count lexicographical order than the first$s_1$Small number string, and then calculates the ratio of lexicographically$s_2$The number of small string, then the two results of a minus, is the answer.
  The first number is better requirements, obtained directly$s_1$It is the first of several aligned itself can be. and$s_3$Because not by$s_1$Converted from a bit harder, so I thought I could be$s_3$Converted into$s_1$A form-letter, and then by the same method requirements.
  In the first of several aligned itself seeking process, with$i$Traversal$0$To$len$Sequentially obtaining: from the comparison results, the current position, is less than the number of strings lexicographically$s_1$. Then for each$i$To determine the results of the current position (here used the permutations knowledge + a small optimization). Finally, adding the answer.

(Formula too lazy to write .. This editor will use a little less. Words, or to knock title, although this topic and ideas out, knock thief slow ...)

Code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long

using namespace std;
const int maxn = 1000010;
const LL mod = 1e9 + 7;

bool s2_smaller, s2_bigger;
int num[30], sum[30];
char s1[maxn], s2[maxn];
LL fac[maxn], inv_fac[maxn];

LL pow(LL a, LL b) {
    LL ret = 1;
    while(b) {
        if(b&1)
            ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}

void pre_treat(int len) {
    fac[0] = inv_fac[0] = 1;
    for(int i=1; i<=len; i++) {
        fac[i] = (LL)i * fac[i-1] % mod;
        inv_fac[i] = pow((LL)fac[i], mod-2);
    }
}

void init(int len) {
    for(int i=0; i<len; i++)
        num[s1[i]-'a'] ++;
    sum[0] = num[0];
    for(int i=1; i<26; i++)
        sum[i] = sum[i-1] + num[i];
}

void transform_s2(int len) {
    init(len);
    for(int i=0; i<len; i++) {
        int alpha = s2[i] - 'a';
        if(!s2_smaller && !s2_bigger) {
            if(num[alpha])
                num[alpha] --;
            else {
                for(int j=alpha-1; j>=0; j--) if(num[j]) {
                    s2_smaller = true; //puts("small");
                    s2[i] = j + 'a'; num[j] --;
                    break ;
                }
                if(s2_smaller) continue ;
                for(int j=alpha+1; j<26; j++) if(num[j]) {
                    s2_bigger = true; //puts("big");
                    s2[i] = j + 'a'; num[j] --;
                    break ;
                }
                if(s2_bigger) continue ;
            }
        }
        else if(s2_smaller) {
            for(int j=25; j>=0; j--) if(num[j]) {
                s2[i] = j + 'a'; num[j] --;
                break ;
            }
        }
        else {//s2_bigger
            for(int j=0; j<26; j++) if(num[j]) {
                s2[i] = j + 'a'; num[j] --;
                break ;
            }
        }
    }
}

LL solve(char *s, int len) {
    init(len);
    LL ret = 0;
    for(int i=0; i<len; i++) {
        int alpha = s[i] - 'a';
        if(alpha > 0 && sum[alpha-1] > 0) {
            LL t1 = fac[len-i-1], t2 = 0;
            for(int j=0; j<26; j++) if(num[j]) {
                t1 = t1 * inv_fac[num[j]] % mod;
                if(j < alpha)
                    t2 = (t2 + num[j]) % mod;
            }
            ret = (ret + t1 * t2 % mod) % mod;
        }
        //printf("------i:%d ret:%I64d\n",i,ret);
        num[alpha] --;
        for(int j=alpha; j<26; j++)
            sum[j] --;
    }
    return ret;
}

int main() {
    //freopen("test.txt","r",stdin);
    s2_smaller = s2_bigger = false;
    scanf("%s%s",s1,s2);
    int len = strlen(s1);
    pre_treat(len);
    transform_s2(len);
    LL l = solve(s1, len), r = solve(s2, len);
    //printf("l:%I64d r:%I64d\n",l,r);
    LL ans = (r - l + mod) % mod;
    if(!s2_smaller && !s2_bigger)
        ans = (ans - 1 + mod) % mod;
    else if(s2_smaller)
        ans = ans;
    else
        ans = (ans - 1 + mod) % mod;
    printf("%I64d\n",ans);
    return 0;
}