Codeforces1575 D. Divisible by Twenty-Five（dp）

Posted 2021-11-05 live4m

题意：

解法：

dp[i][j][0/1]表示前i个数,%25=j,最高位是否有非0位(涉及前导零),的方案数.

特判n=1的情况.

如果无X,那么直接dp即可.
如果有X,那么O(10)枚举X,然后dp累加答案即可.

code：

#include<bits/stdc++.h>
// #define SYNC_OFF
typedef std::vector<int> VE;
typedef std::pair<int,int> PI;
#define int long long
#define ll long long
#define ull unsigned long long
//fast-coding
#define all(a,n) a+1,a+1+n
#define ff(i,n) for(ll i=1;i<=n;i++)
#define rff(i,n) for(ll i=n;i>=1;i--)
#define fff(i,n) for(ll i=0;i<n;i++)
#define rfff(i,n) for(ll i=n-1;i>=0;i--)
#define SC(x) scanf("%s",x)
#define SL(x) strlen(x)
#define pss(a) push_back(a)
#define ps(a) push(a)
#define SZ(x) (int)x.size()
#define pee puts("");
#define eee putchar(' ');
#define re readdd()
#define pr(a) printtt(a)
int readdd(){int x=0,f=1;char c=getchar();//
while(!isdigit(c)&&c!='-')c=getchar();
if(c=='-')f=-1,c=getchar();
while(isdigit(c))x=x*10+c-'0',c=getchar();
return f*x;}
void printtt(int x){if(x<0)putchar('-'),x=-x;//
if(x>=10)printtt(x/10);putchar(x%10+'0');}
int gcd(int a,int b){return b==0?a:gcd(b,a%b);}//
int ppow(int a,int b,int mod){a%=mod;//
int ans=1%mod;while(b){if(b&1)ans=(long long)ans*a%mod;
a=(long long)a*a%mod;b>>=1;}return ans;}
bool addd(int a,int b){return a>b;}
int lowbit(int x){return x&-x;}
const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};
bool isdigit(char c){return c>='0'&&c<='9';}
bool Isprime(int x){
    for(int i=2;i*i<=x;i++)if(x%i==0)return 0;
    return 1;
}
void ac(int x){if(x)puts("YES");else puts("NO");}
//
using namespace std;
// const int mod=998244353;
const int mod=1e9+7;
const int maxm=2e6+5;
char s[10];
int d[10][25][2];
int n;
void dp(){
    for(int i=0;i<=n;i++){
        for(int j=0;j<25;j++){
            for(int f=0;f<2;f++){
                d[i][j][f]=0;
            }
        }
    }
    d[0][0][0]=1;
    for(int i=0;i<n;i++){
        for(int j=0;j<25;j++){
            for(int f=0;f<2;f++){
                if(!d[i][j][f])continue;
                if(s[i+1]=='_'){
                    for(int x=0;x<10;x++){
                        if(!f&&!x)continue;
                        int nf=(f||x);
                        d[i+1][(j*10+x)%25][nf]+=d[i][j][f];
                    }
                }else{
                    int x=s[i+1]-'0';
                    if(!f&&!x)continue;
                    int nf=(f||x);
                    d[i+1][(j*10+x)%25][nf]+=d[i][j][f];
                }
            }
        }
    }
}
void solve(){
    SC(s+1);
    n=SL(s+1);
    vector<int>pos;
    ff(i,n)if(s[i]=='X')pos.pss(i);
    if(n==1){
        if(pos.size()||s[1]=='0'||s[1]=='_'){
            pr(1);pee;
        }else{
            pr(0);pee;
        }
        return ;
    }
    //
    int ans=0;
    if(pos.size()){
        for(int i=0;i<10;i++){
            for(auto x:pos){
                s[x]=i+'0';
            }
            dp();
            ans+=d[n][0][0]+d[n][0][1];
        }
    }else{
        dp();
        ans+=d[n][0][0]+d[n][0][1];
    }
    //
    pr(ans);pee;
}
void Main(){
    // #define MULTI_CASE
    #ifdef MULTI_CASE
    int T;cin>>T;while(T--)
    #endif
    solve();
}
void Init(){
    #ifdef SYNC_OFF
    ios::sync_with_stdio(0);cin.tie(0);
    #endif
    #ifndef ONLINE_JUDGE
    freopen("../in.txt","r",stdin);
    freopen("../out.txt","w",stdout);
    #endif
}
signed main(){
    Init();
    Main();
    return 0;
}

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