2019-2020 ICPC Asia Hong Kong Regional Contest J. Junior Mathematician 题解（数位dp）

题目链接

题目大意

要你在[l,r]中找到有多少个数满足(xequiv f(x)(mod; m))

(f(x)=sum_{i=1}^{k-1} sum_{j=i+1}^{k}d(x,i)*d(x,j))

(d(x,i)表示x的第i位数)

题目思路

显然是数位dp，然而这个数位dp不能同时存x%m 和f(x)%m

这样会内存太大存不了，所以存差值即可

还有这个dfs的时候取模只取一次，不然会t，卡常严重

代码

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#define fi first
#define se second
#define debug printf(" I am here
");
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=5e3+5,inf=0x3f3f3f3f,mod=1e9+7;
const double eps=1e-10;
char l[maxn],r[maxn];
int m,base[maxn];
ll dp[maxn][100][100];
int lenl,lenr,num[maxn];
ll dfs(int pos,int pre,int dif,bool flag){
    if(pos==0) return dif==0;
    if(!flag&&dp[pos][pre][dif]!=-1){
        return dp[pos][pre][dif];
    }
    int lim=flag?num[pos]:9;
    ll ans=0;
    for(int i=0;i<=lim;i++){
        ans=ans+dfs(pos-1,(pre+i)%m,((dif+i*base[pos]-i*pre)%m+m)%m,flag&&i==lim);
    }
    ans%=mod;
    if(!flag){
        dp[pos][pre][dif]=ans;
    }
    return ans;
}
ll solve1(){
    for(int i=1;i<=lenr;i++){
        num[i]=r[lenr-i+1]-'0';
    }
    return dfs(lenr,0,0,1);
}
ll solve2(){
    for(int i=1;i<=lenl;i++){
        num[i]=l[lenl-i+1]-'0';
    }
    return dfs(lenl,0,0,1);
}
bool check(){
    int sum1=0,sum2=0,pre=0;
    for(int i=1;i<=lenl;i++){
        sum1=(sum1*10+l[i]-'0')%m;//x%m
        sum2=(sum2+(l[i]-'0')*pre)%m;// f(x)%m
        pre=(pre+l[i]-'0')%m;
    }
    return sum1==sum2;
}
signed main(){
    int _;scanf("%d",&_);
    while(_--){
        scanf("%s%s",l+1,r+1);
        scanf("%d",&m);
        lenl=strlen(l+1),lenr=strlen(r+1);
        for(int i=1;i<=max(lenl,lenr);i++){
            for(int j=0;j<m;j++){
                for(int k=0;k<=m;k++){
                    dp[i][j][k]=-1;
                }
            }
        }
        base[1]=1;
        for(int i=2;i<=max(lenl,lenr);i++){
            base[i]=base[i-1]*10%m;
        }
        ll ans=((solve1()-solve2()+check())%mod+mod)%mod;
        printf("%lld
",ans);
    }
    return 0;
}