【HDOJ】4351 Digital root

digital root = n==0 ? 0 : n%9==0 ? 9:n%9;
可以简单证明一下
n = a0*n^0 + a1*n^1 + ... + ak * n^k
n%9 = a0+a1+..+ak
然后，数学归纳易知结论是正确的。
因此9个状态就够了，表示%9的结果。
这里需要特殊处理0， 表示状态为0。

/* 4351 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000")

#define sti                set<int>
#define stpii            set<pair<int, int> >
#define mpii            map<int,int>
#define vi                vector<int>
#define pii                pair<int,int>
#define vpii            vector<pair<int,int> >
#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define clr                clear
#define pb                 push_back
#define mp                 make_pair
#define fir                first
#define sec                second
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1

typedef struct {
    short sum;
    short lst, rst, st;
} node_t;

const int maxn = 1e5+5;
int zn[maxn];
bool visit[10];
int a[maxn];
node_t nd[maxn<<2];
int Mod9[105];
short sum_st[20][512];
short st_st[512][512];
int n;

void init() {
    rep(i, 0, 105)
        Mod9[i] = i % 9;
        
    rep(st, 0, 512) {
        rep(i, 0, 20) {
            int nst = 0;
            
            rep(j, 0, 9) {
                if (st & (1<<j)) {
                    int k = Mod9[i + j];
                    nst |= (1<<k);
                }
            }
            sum_st[i][st] = nst;
        }
    }
    
    rep(lst, 0, 512) {
        rep(rst, lst, 512) {
            int nst = 0;
            rep(i, 0, 9) {
                if ((lst & (1<<i)) == 0)
                    continue;
                rep(j, 0, 9) {
                    if (rst & (1<<j)) {
                        int k = Mod9[i+j];
                        nst |= (1<<k);
                    }
                }
            }
            st_st[lst][rst] = st_st[rst][lst] = nst;
        }
    }
}


void PushUp(int rt) {
    int lb = rt << 1;
    int rb = lb | 1;

    nd[rt].sum = Mod9[nd[lb].sum + nd[rb].sum];
    nd[rt].lst = nd[lb].lst | sum_st[nd[lb].sum][nd[rb].lst];
    nd[rt].rst = nd[rb].rst | sum_st[nd[rb].sum][nd[lb].rst];
    nd[rt].st = nd[lb].st | nd[rb].st | st_st[nd[lb].rst][nd[rb].lst];
}

void Build(int l, int r, int rt) {
    if (l == r) {
        if (zn[l] == zn[l-1]) {
            nd[rt].sum = a[l];
            nd[rt].lst = nd[rt].rst = nd[rt].st = (1 << a[l]);
        } else {
            nd[rt].sum = a[l];
            nd[rt].lst = nd[rt].rst = nd[rt].st = 0;
        }
        return ;
    }

    int mid = (l + r) >> 1;

    Build(lson);
    Build(rson);

    PushUp(rt);
}

node_t Query(int L, int R, int l, int r, int rt) {
    if (L==l && R==r) {
        return nd[rt];
    }

    int mid = (l + r) >> 1;

    if (R <= mid) {
        return Query(L, R, lson);
    } else if (L > mid) {
        return Query(L, R, rson);
    } else {
        node_t lnd = Query(L, mid, lson);
        node_t rnd = Query(mid+1, R, rson);
        node_t ret;

        ret.sum = Mod9[lnd.sum + rnd.sum];
        ret.lst = lnd.lst | sum_st[lnd.sum][rnd.lst];
        ret.rst = rnd.rst | sum_st[rnd.sum][lnd.rst];
        ret.st = lnd.st | rnd.st | st_st[lnd.rst][rnd.lst];
        
        return ret;
    }
}

void solve() {
    Build(1, n, 1);

    int q;
    node_t d;
    int l, r;

    scanf("%d", &q);
    while (q--) {
        scanf("%d %d", &l, &r);
        memset(visit, false, sizeof(visit));
        d = Query(l, r, 1, n, 1);
        rep(i, 0, 9) {
            if (d.st & (1<<i))
                visit[i] = true;
        }
        visit[9] = visit[0];
        visit[0] = (zn[r] - zn[l-1]) > 0;
        int cnt = 5;
        per(i, 0, 10) {
            if (visit[i]) {
                if (cnt == 5)
                    printf("%d", i);
                else
                    printf(" %d", i);
                if (--cnt == 0)
                    break;
            }
        }

        while (cnt) {
            printf(" -1");
            --cnt;
        }
        putchar('
');
    }
}

int main() {
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
        freopen("data.out", "w", stdout);
    #endif

    int t;

    init();
    scanf("%d", &t);
    rep(tt, 1, t+1) {
        scanf("%d", &n);
        rep(i, 1, n+1) {
            scanf("%d", &a[i]);
            zn[i] = zn[i-1] + (a[i] == 0);
            a[i] %= 9;
        }
        printf("Case #%d:
", tt);
        solve();
        if (tt != t)
            putchar('
');
    }

    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif

    return 0;
}

数据发生器。

from copy import deepcopy
from random import randint, shuffle
import shutil
import string


def GenDataIn():
    with open("data.in", "w") as fout:
        t = 10
        bound = 10**9
        fout.write("%d
" % (t))
        for tt in xrange(t):
            n = randint(10, 20)
            fout.write("%d
" % (n))
            L = []
            for i in xrange(1, n):
                x = randint(0, bound)
                L.append(x)
            L.append(0)
            shuffle(L)
            fout.write(" ".join(map(str, L)) + "
")
            q = (n+1)*n/2
            fout.write("%d
" % (q))
            for l in xrange(1, n+1):
                for r in xrange(l, n+1):
                    fout.write("%d %d
" % (l, r))
                
                
def MovDataIn():
    desFileName = "F:eclipse_prjworkspacehdojdata.in"
    shutil.copyfile("data.in", desFileName)

    
if __name__ == "__main__":
    GenDataIn()
    MovDataIn()