SPOJ BALNUM Balanced Numbers（数位DP+状态压缩）题解

思路：

把0~9的状态用3进制表示，数据量3^10

代码：

#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define ll long long
const int N = 500+5;
const int INF = 0x3f3f3f3f;
using namespace std;
int dp[22][60000];  //0 没出现，1 奇数，2 偶数
int dig[22];
int check(int sta){
    for(int i = 0;i <= 9;i++){
        int tmp = sta % 3;
        sta /= 3;
        if(i % 2 == 1 && tmp == 1) return 0;
        else if(i % 2 == 0 && tmp == 2) return 0;
    }
    return 1;
}
int news(int sta,int x){
    int num[10];
    for(int i = 0;i <= 9;i++){
        num[i] = sta % 3;
        sta /= 3;
    }
    int tmp = 0;
    if(num[x] == 0) num[x] = 1;
    else num[x] = 3 - num[x];
    for(int i = 9;i >= 0;i--){
        tmp = tmp*3 + num[i];
    }
    return tmp;
}
ll dfs(int pos,int sta,bool lead,bool limit){
    if(pos == -1) return check(sta);
    if(!limit && dp[pos][sta] != -1) return dp[pos][sta];
    int top = limit? dig[pos] : 9;
    ll ret = 0;
    for(int i = 0;i <= top;i++){
        int STA = news(sta,i);
        if(lead && i == 0) STA = 0;
        ret += dfs(pos - 1,STA,lead && i == 0,limit && i == top);
    }
    if(!limit) dp[pos][sta] = ret;
    return ret;
}
ll solve(ll x){
    int pos = 0;
    if(x == 0) return 1;
    while(x){
        dig[pos++] = x % 10;
        x /= 10;
    }
    return dfs(pos - 1,0,true,true);
}
int main(){
    int T;
    ll a,b;
    scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
    while(T--){
        scanf("%lld%lld",&a,&b);
        printf("%lld\n",solve(b) - solve(a - 1));
    }
    return 0;
}