SPOJ BALNUM Balanced Numbers （数位dp）-优快云博客

本文链接：https://blog.youkuaiyun.com/lwt36/article/details/49317513

本文介绍了一种使用三进制压缩的动态规划算法解决特定数值区间内数字特征计数问题的方法。通过将数字特性（奇数位和偶数位的数量）转化为三进制状态进行压缩，有效减少了状态空间。

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题意:

$求[l,r]区间中数满足, 奇数数字有偶数个, 偶数数字有奇数个, 的个数$

分析:

$三进制压缩 - - sta[0\to 9]:= 0没有出现, 1出现奇数次, 2出现偶数次$
$把sta三进制压缩, dp[i][s]:= i位, 状态s, (s \leq 3^{10} = 59049)$
$转移什么的都很显然, 同样注意前导0, 10^{19}爆LL了, 开ULL$

代码:

//
//  Created by TaoSama on 2015-10-21
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#include <climits>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

typedef unsigned long long ULL;

ULL dp[22][59055]; //3^10 = 59049
int digit[22], sta[10], vis[22][59055], kase;
//0 for none, 1 for odd, 2 for even

int encode() {
    int s = 0;
    for(int i = 0; i < 10; ++i) s = s * 3 + sta[i];
    return s;
}

void decode(int s) {
    for(int i = 9; ~i; --i, s /= 3) sta[i] = s % 3;
}

void print() {
    for(int i = 0; i < 10; ++i) cout << sta[i] << ' '; cout << endl;
}

bool check(int s) {
    decode(s);
    for(int i = 0; i < 10; ++i) {
        if((i & 1) && sta[i] == 1) return false;
        if(!(i & 1) && sta[i] == 2) return false;
    }
    return true;
}

int trans(int s, int d) {
    decode(s);
    if(sta[d] == 0) sta[d] = 1;
    else sta[d] = 3 - sta[d];
    return encode();
}

ULL dfs(int i, int s, bool first, bool e) {
    if(!i) return check(s);
    if(!e && vis[i][s] == kase) return dp[i][s];
    int to = e ? digit[i] : 9;
    ULL ret = 0;
    for(int d = 0; d <= to; ++d)
        ret += dfs(i - 1, first && !d ? s : trans(s, d),
                   first && !d, e && d == to);
    if(!e) vis[i][s] = kase, dp[i][s] = ret;
    return ret;
}

ULL calc(ULL x) {
    int cnt = 0;
    for(; x; x /= 10) digit[++cnt] = x % 10;
    return dfs(cnt, 0, 1, 1);
}

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int t; scanf("%d", &t);
    while(t--) {
        ULL l, r; scanf("%llu%llu", &l, &r);
        ++kase;
        printf("%llu\n", calc(r) - calc(l - 1));
    }
    return 0;
}