SPOJ BALNUM Balanced Numbers （数位dp）

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→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≤310=59049)
转移什么的都很显然,同样注意前导0,1019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 << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef unsigned long long ULL;ULL dp[22][59055]; //3^10 = 59049int digit[22], sta[10], vis[22][59055], kase;//0 for none, 1 for odd, 2 for evenint 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;}