洛谷 P3413 SAC#1

题目传送门

数位dp.

#include<iostream>
#include<cstdio>
#include<cstring>
#define mod 1000000007 

using namespace std;

string l,r;
int a[1005],b[1005],f[1005][15][15][2][2][2];

inline long long solve1(int d,int pr,int prp,bool li,bool qd,bool s)  {
    long long sum = 0;
    if(d == l.length() + 1) return s;
    if(f[d][pr][prp][li][qd][s] != -1) return f[d][pr][prp][li][qd][s];
    int res = li ? a[d] : 9;
    for(int i = 0;i <= res; i++)
        sum = (sum + solve1(d + 1,(qd && i == 0) ? -1 : i,pr,li && (i == res),qd && (i == 0),s || (i == pr && pr != -1 || i == prp && pr != -1 && prp != -1)) % mod) % mod;
    f[d][pr][prp][li][qd][s] =  sum;
    return sum;
}

inline long long solve2(int d,int pr,int prp,bool li,bool qd,bool s)  {
    long long sum = 0;
    if(d == r.length() + 1) return s;
    if(f[d][pr][prp][li][qd][s] != -1) return f[d][pr][prp][li][qd][s];
    int res = li ? b[d] : 9;
    for(int i = 0;i <= res; i++)
        sum = (sum + solve2(d + 1,(qd && i == 0) ? -1 : i,pr,li && (i == res),qd && (i == 0),s || (i == pr && pr != -1 || i == prp && pr != -1 && prp != -1)) % mod) % mod;
    f[d][pr][prp][li][qd][s] =  sum;
    return sum;
}

int main() {
    cin >> l >> r;
    for(int i = 0;i < l.length(); i++)    
        a[i+1] = l[i] - '0';
    for(int i = 0;i < r.length(); i++)
        b[i+1] = r[i] - '0';
    a[l.length()]--;
    memset(f,-1,sizeof(f));
    long long ans1 = solve1(1,-1,-1,1,1,0) % mod;
    memset(f,-1,sizeof(f));
    long long ans2 = solve2(1,-1,-1,1,1,0) % mod;
    printf("%d",(ans2 - ans1 + mod) % mod);
    return 0;
} 

#include<iostream>
#include<cstdio>
#include<cstring>
#define mod 1000000007

using namespace std;

int t,s[20],len;
long long l,r,ans,ans1,f[20][2][15];

inline void fenjie(long long x) {
    len = 0;
    while(x) {
        s[++len] = x % 10;
        x /= 10;
    }
}

inline long long dfs(int d,int sum,bool li,int p) {
    if(d == 0) return sum % mod;
    if(f[d][li][p] != -1) return f[d][li][p];
    long long ss = 0;
    int res = li ? s[d] : 9;
    for(int i = 0;i <= res; i++)
        ss = (ss + dfs(d - 1,sum + (i == p),li && (i == res),p) % mod) % mod;
    f[d][li][p] = ss % mod;
    return ss % mod;
}

inline void chushihua() {
    ans = ans1 = 0;
    len = 0;
    memset(f,-1,sizeof(f));
    memset(s,0,sizeof(s));
}

int main() {
    scanf("%d",&t);
    while(t--) {
        scanf("%lld%lld",&l,&r);
        chushihua();
        fenjie(l - 1);
        for(int i = 1;i <= 9; i++) {
            memset(f,-1,sizeof(f));
            ans = (ans + (dfs(len,0,1,i) % mod * i) % mod) % mod;
        }
        fenjie(r);
        for(int i = 1;i <= 9; i++) {
            memset(f,-1,sizeof(f));
            ans1 = (ans1 + (dfs(len,0,1,i) % mod * i) % mod) % mod;
        }
        printf("%d
",(ans1 - ans + mod) % mod);
    }
    return 0;
}

AC代码1是自己写的,代码比较复杂.

AC代码2是自己写后照着题解改的,代码比较简洁