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「HDU2089」不要62 数位dp

发表于 | 分类于
字数统计: 475字 | 阅读时长 ≈ 3分钟

Links there:HDU-2089-不要62

题意

身为杭州人看到这个题意笑出了声.

求规定区间内数字中满足不含62连号、不含4的数字.

思路

和windy数差不多的 反正都是一个套路.

预处理之后再统计即可

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//my vegetable has exploded. :(
#include<bits/stdc++.h>
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MM(x,y) memset(x,y,sizeof(x))
#define MCPY(a,b) memcpy(a,b,sizeof(b))
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,b,a) for(int i=b;i>=a;i--)
#define fi first
#define se second
using namespace std;
#define int long long

inline int quickpow(int m,int n,int p){int b=1;while(n){if(n&1)b=b*m%p;n=n>>1;m=m*m%p;}return b;}
inline int getinv(int x,int p){return quickpow(x,p-2,p);}
inline int read(void){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){f=ch=='-'?-1:1;ch=getchar();}
    while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x * f;
}

int f[14][14];
int a[14],cnt=0;

///------------------head------------------
inline void init(void){
    rep(i,0,9) f[1][i] = 1;
    f[1][4] = 0;
    rep(i,2,13)
        rep(j,0,9)
        {
            if (j == 4) continue;
            rep(k,0,9)
            {
                if ((k == 2 && j == 6)) continue;
                f[i][j] += f[i-1][k];
            }
        }
}

inline int calc(int x){
    int ret = 0; cnt = 0;
    while(x){a[++cnt]=x%10;x/=10;}
    rep(i,1,cnt-1) rep(j,1,9) ret += f[i][j];
    rep(j,1,a[cnt]-1) ret += f[cnt][j];
    per(i,cnt-1,1)
    {
        rep(j,0,a[i]-1)
        if (j != 4 && a[i+1] != 4)
        ret += f[i][j] * (!(a[i+1] == 6 && j == 2));
        if ((a[i+1] == 6 && a[i] == 2) || a[i+1] == 4 || a[i] == 4) break;
        if (i==1) ++ret;
    }
    return ret;
}

signed main(signed argc, char *argv[]){
    init();
    int a,b;
    while(scanf("%lld %lld",&a,&b) == 2 && (a+b)) printf("%lld\n",calc(b)-calc(a-1)+((!a)||(!b)));
    return 0;
}