hdu3652 && hdu 4722 && hdu3555

标签：

数位dp简单题

hdu3652

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int N = 10+10;
int dig[N];
ll dp[N][N][2][20];
ll dfs(int len,int pre,int is,int sum,int f)
{
    if(len < 1) return is&&(sum==0) ;
    if(!f && dp[len][pre][is][sum]!=-1) return dp[len][pre][is][sum];
    int last = f ? dig[len] : 9;
    ll res = 0;
    for(int i=0;i<=last;i++)
    {
        res += dfs(len-1,i,is||(pre==1&&i==3),(sum*10+i)%13,f&&(i==last) );
    }
    if(!f) dp[len][pre][is][sum] = res;
    return res;
}
ll solve(int n)
{
    int len = 0;
    while(n)
    {
        dig[++len] = n % 10;
        n /= 10;
    }
    return dfs(len,0,0,0,1);
}
int main()
{
    int t,n;
    ll ans;
    memset(dp,-1,sizeof(dp));
    while(scanf("%d",&n)!=EOF)
    {
        ans = solve(n);
        printf("%lld\n",ans);
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 20+5;
int dig[N];
ll dp[N][20];
ll dfs(int len ,int sum,int f)
{
    if(len<1) return sum%10==0;
    if(!f && dp[len][sum]!=-1) return dp[len][sum];
    int last = f ? dig[len] : 9;
    ll res = 0;
    for(int i=0;i<=last;i++)
    {
        res += dfs(len-1,(sum+i)%10,f&&(i==last));
    }
    if(!f) dp[len][sum] = res;
    return res;
}
ll solve(ll n)
{
    if(n<0) return 0;
    int len = 0;
    while(n)
    {
        dig[++len] = n%10;
        n/=10;
    }
    return dfs(len,0,1);
}
int main()
{
    int t;
    int cas = 0;
    ll ans,l,r;
    cin>>t;
    memset(dp,-1,sizeof(dp));
    while(t--)
    {
        scanf("%lld%lld",&l,&r);
        ans = solve(r) - solve(l-1);
        printf("Case #%d: %lld\n",++cas,ans);
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 20+5;
int dig[N];
ll dp[N][10][2];
ll dfs(int len ,int pre,int is,int f)
{
    if(len<1) return is;
    if(!f && dp[len][pre][is]!=-1) return dp[len][pre][is];
    int last = f ? dig[len] : 9;
    ll res = 0;
    for(int i=0;i<=last;i++)
    {
        res += dfs(len-1,i,is||(pre==4&&i==9),f&&(i==last));
    }
    if(!f) dp[len][pre][is] = res;
    return res;
}
ll solve(ll n)
{
    int len = 0;
    while(n)
    {
        dig[++len] = n%10;
        n/=10;
    }
    return dfs(len,0,0,1);
}
int main()
{
    int t;
    ll ans,r;
    cin>>t;
    memset(dp,-1,sizeof(dp));
    while(t--)
    {
        scanf("%lld",&r);
        ans = solve(r);
        printf("%lld\n",ans);
    }
    return 0;
}

hdu3652 && hdu 4722 && hdu3555

标签：

原文地址：http://blog.csdn.net/alpc_wt/article/details/44043123