hud 2089 不要62 hdu 3555 Bomb

这是碰到的第一类数位dp：找这个数是否存在子串为一个一位数，或两位数

其实都一样，记录已经满足前几位即可。

特定的：

如果前面计算对后面有影响，计算前驱的值ten[]

hud 2089

初学，根据我自己的理解码代码，无限wa，不知道是错在哪里，后来强行手跑代码，发现先前的理解不准确。

数位dp的记忆化搜索不能理解为边爆搜边记录，这样是没有意义的（之前脑子懵）。

它的搜索只是搜了（0000~0009），记录到dp数组，然后根据其搜10次（001X~009X），以此类推。

wa代码

#include<iostream>#include<cstdio>#include<cstring>#include<stack>#include<map>#include<queue>#include<cmath>#include<algorithm>#include<deque>typedef long long LL;using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")const int INF=0x3f3f3f3f;const int N = 20;const double eps = 1e-6;int dp[N]; //0->10^i-1int ten[N],f[N];void init(){    ten[0] = 1;    for(int i = 1; i < 10; i++)        ten[i] = ten[i-1]*10;}int dfs(int pos,int num,int limit,int pre){    if(pos == -1)        return num;    if(!limit && dp[pos] != -1)    {        if(num) return ten[pos+1];        else    return dp[pos+1];    }    int top = limit ? f[pos]:9;    //printf("pos = %d top = %d\n",pos,top);    int ans = 0;    for(int i = 0; i <= top; i++)        ans += dfs(pos-1,num||(i==4)||(i==2&&pre==6),limit && i==f[pos],i);    if(!limit)        dp[pos+1] = ans;    return ans;}int main(){    init();    int a,b;    while(~scanf("%d%d",&a,&b) && a,b)    {        memset(dp,-1,sizeof(dp));        int pos = 0;        int aa = a-1;        while(aa)        {            f[pos++] = aa%10;            aa /= 10;        }        int suma = dfs(pos-1,0,1,0);        //printf("%d\n",suma);        memset(dp,-1,sizeof(dp));        pos = 0;        int bb = b;        while(bb)        {            f[pos++] = bb%10;            bb /= 10;        }        int sumb = dfs(pos-1,0,1,0);        printf("%d\n",b-a+1-(sumb-suma));    }    return 0;}

ac代码

#include<iostream>#include<cstdio>#include<cstring>#include<stack>#include<map>#include<queue>#include<cmath>#include<algorithm>#include<deque>typedef long long LL;using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")const int INF=0x3f3f3f3f;const int N = 20;const double eps = 1e-6;int dp[N][3]; //0->10^i-1 dp[i][2]表示有4或有62的个数,dp[i][1]表示首位有6的个数,dp[i][0]表示没有任何不吉利特征的个数int ten[N],f[N];void init(){    ten[0] = 1;    for(int i = 1; i < 10; i++)        ten[i] = ten[i-1]*10;}int dfs(int pos,int num,int limit){    if(pos == 0)        return num==2;    if(!limit && dp[pos][num] != -1)        return dp[pos][num];    int top = limit ? f[pos]:9;    int ans = 0;    for(int i = 0; i <= top; i++)    {        if(num == 2 || (num==1&&i==2) || i==4)            ans += dfs(pos-1,2,limit&&i==f[pos]);        else if(i == 6)            ans += dfs(pos-1,1,limit&&i==f[pos]);        else            ans += dfs(pos-1,0,limit&&i==f[pos]);    }    if(!limit)        dp[pos][num] = ans;    return ans;}int solve(int a){    int pos = 0;    while(a)    {        f[++pos] = a%10;        a /= 10;    }    return dfs(pos,0,1);}int main(){    init();    int a,b;    memset(dp,-1,sizeof(dp));    while(~scanf("%d%d",&a,&b) && a,b)    {        int suma = solve(a-1);        int sumb = solve(b);        printf("%d\n",b-a+1-(sumb-suma));    }    return 0;}

同一类型 hud 3555

#include<iostream>#include<cstdio>#include<cstring>#include<stack>#include<map>#include<queue>#include<cmath>#include<algorithm>#include<deque>typedef long long LL;using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")const int INF=0x3f3f3f3f;const int N = 25;const double eps = 1e-6;LL dp[N][3]; //0->10^i-1   dp[i][2]表示49的个数,dp[i][1]表示首位是4的个数,dp[i][0]表示不包含49和4的个数LL ten[N];int f[N];//void init()//{//    ten[0] = 1;//    for(int i = 1; i < N; i++)//        ten[i] = ten[i-1]*10;//}LL dfs(int pos,int zt,int limit){    if(pos == 0)        return zt==2;    if(!limit && dp[pos][zt] != -1) {        return dp[pos][zt];    }    int top = limit ? f[pos]:9;    LL ans = 0;    for(int i = 0; i <= top; i++)    {        if(zt==2 || (zt==1&&i==9))            ans += dfs(pos-1,2,limit && i==f[pos]);        else if(i==4)            ans += dfs(pos-1,1,limit && i==f[pos]);        else            ans += dfs(pos-1,0,limit && i==f[pos]);    }    if(!limit)        dp[pos][zt] = ans;    return ans;}LL solve(LL a){    int pos = 0;    while(a)    {        f[++pos] = a%10;        a /= 10;    }    return dfs(pos,0,1);}int main(){    int T;    scanf("%d",&T);    memset(dp,-1,sizeof(dp));    while(T--)    {        LL n;        scanf("%I64d",&n);        printf("%I64d\n",solve(n));    }    return 0;}