hdu_2089(数位dp)

hdu_2089(数位dp)

标签： dp

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我初次接触数位dp表面上看上去挺简单，但是仔细学还是要考虑很多细节的。wa了无数次，这里引入一个
很好地博客

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
ll dp[11][11];

void init()
{
    dp[0][0] = 1;
    for(int i = 1; i < 7; i++){
        for(int j = 0; j <= 9; j++){
            if(j==4) continue;
            for(int k = 0; k <= 9; k++){
                if(k!=4&&!(j==6&&k==2)){
                    dp[i][j] += dp[i-1][k];
                }
            }
        }
    }
}

int a[10];

ll solve(int n){
    ll ans = 0;
    int len = 0;
    while(n){
        a[len++] = n%10;
        n = n/10;
    }
    int last = 0;
    a[len] = 0;//防止特判只有一位的情况
    for(int i = len-1; i >= 0; i--){
        for(int j = 0; j < a[i]; j++){
          /*  if(j!=4 && !(j==2&&last==6)){
                ans += dp[i+1][j];
            }
            */
            if(j==4 || (last==6 && j == 2))continue;
            ans += dp[i+1][j];
        }
        if(a[i]==4 || (a[i]==2&&last==6)) break;
        last = a[i];
    }
    return ans;
}

int main()
{
    init();
    ll n,m;
    while(~scanf("%I64d %I64d",&n,&m))
    {
        if(m==0&&n==0) return 0;
        ll ans = solve(m+1)-solve(n);
        printf("%I64d\n",ans);
    }
    return 0;
}

hdu_3555(数位dp)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
int a[30];
ll dp[30][3];
void init()
{
    dp[0][0] = 1;
    dp[0][1] = dp[0][2] = 0;
    for(int i = 1; i < 30; i++)
    {
        dp[i][0] = dp[i-1][0]*10 - dp[i-1][1];
        dp[i][1] = dp[i-1][0];
        dp[i][2] = dp[i-1][1] + dp[i-1][2]*10;
    }
}
int main()
{
    init();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll n;
        scanf("%I64d",&n);
        n = n+1;//
        int len = 0;
        while(n)
        {
            a[len++] = n%10;
            //printf("a...%d ",a[len-1]);
            n/=10;
        }
       // printf("%d\n",len);
        bool fl = 0;
        ll ans = 0;
        int last = 0;
        for(int i = len-1; i >= 0; i--){
            ans+=dp[i][2]*a[i];
            if(fl){
                ans += dp[i][0]*a[i];
            }
            else if(a[i]>4) ans+=dp[i][1];
            if(last == 4 &&a[i] == 9)fl = 1;
            last = a[i];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}