【题解】Willem, Chtholly and Seniorious Codeforces 896C ODT

Prelude

ODT这个东西真是太好用了,以后写暴力骗分可以用,写在这里mark一下。
题目链接:ヽ(✿゚▽゚)ノ


Solution

先把原题解贴在这里:(ノ*・ω・)ノ
简单地说,因为数据是全部随机的,所以一定会有特别多的区间set,就会有很多数字相同,那么我们暴力把相同的数字合并成一个点,合并完之后数组就变得很短,然后对于询问暴力做就可以了。
具体复杂度是什么我也不会证明qwq,所以就把由乃的题解贴上来了qwq,感觉很靠谱的样子。
题解中给的是用STL的set维护缩点后的数组,我感觉不是很好写,就直接用数组维护了。
ddd还写过链表维护的,不过感觉都差不多,反正复杂度都是对的,肯定能过qwq。


Code

#include <bits/stdc++.h>
#define dprint printf

using namespace std;
typedef long long ll;
typedef pair<ll,int> pli;
const int MAXN = 100010;
int _w;

void noprint(...) {}

int fpow( int a, int b, int mod ) {
    int c = 1;
    while(b) {
        if( b & 1 ) c = int(1LL * c * a % mod);
        a = int(1LL * a * a % mod);
        b >>= 1;
    }
    return c;
}

int inter( int l1, int r1, int l2, int r2 ) {
    int l = max(l1, l2);
    int r = min(r1, r2);
    return max(r-l+1, 0);
}

int n, m, seed, vmax;
ll a[MAXN];

int rnd() {
    const int MOD = 1e9+7;
    
    int ret = seed;
    seed = int((1LL * seed * 7 + 13) % MOD);
    return ret;
}

pli b[MAXN];
int sz;
void prelude() {
    sz = 1;
    b[sz-1].first = a[1], b[sz-1].second = 0;
    for( int i = 1; i <= n; ++i ) {
        if( b[sz-1].first == a[i] ) {
            ++b[sz-1].second;
        } else {
            b[sz].first = a[i], b[sz].second = 1, ++sz;
        }
    }
}

pli c[MAXN];
int csz;

void append( ll x, int cnt ) {
    if( csz && c[csz-1].first == x )
        c[csz-1].second += cnt;
    else
        c[csz].first = x, c[csz].second = cnt, ++csz;
}

void solve1( int l, int r, int x ) {
    int L = 0, R = 0;
    csz = 0;
    for( int i = 0; i < sz; ++i ) {
        L = R+1;
        R = L + b[i].second - 1;
        ll num = b[i].first;
        int cntl = inter(L, l-1, L, R);
        int cnt = inter(L, R, l, r);
        int cntr = inter(r+1, R, L, R);
        if( cntl ) append(num, cntl);
        if( cnt ) append(num+x, cnt);
        if( cntr ) append(num, cntr);
    }
    sz = csz;
    for( int i = 0; i < sz; ++i )
        b[i] = c[i];
}

void solve2( int l, int r, int x ) {
    int L = 0, R = 0;
    csz = 0;
    for( int i = 0; i < sz; ++i ) {
        L = R+1;
        R = L + b[i].second - 1;
        ll num = b[i].first;
        int cntl = inter(L, l-1, L, R);
        int cnt = inter(L, R, l, r);
        int cntr = inter(r+1, R, L, R);
        if( cntl ) append(num, cntl);
        if( cnt ) append(x, cnt);
        if( cntr ) append(num, cntr);
    }
    sz = csz;
    for( int i = 0; i < sz; ++i )
        b[i] = c[i];
}

void solve3( int l, int r, int x ) {
    int L = 0, R = 0;
    csz = 0;
    for( int i = 0; i < sz; ++i ) {
        L = R+1;
        R = L + b[i].second - 1;
        int cnt = inter(L, R, l, r);
        if( cnt ) c[csz++] = pli( b[i].first, cnt );
    }
    sort(c, c+csz);
    L = R = 0;
    for( int i = 0; i < csz; ++i ) {
        L = R+1;
        R = L + c[i].second - 1;
        if( x >= L && x <= R )
            return (void)printf( "%lld\n", c[i].first );
    }
    return assert(0);
}

void solve4( int l, int r, int x, int y ) {
    int L = 0, R = 0, ans = 0;
    for( int i = 0; i < sz; ++i ) {
        L = R+1;
        R = L + b[i].second - 1;
        int cnt = inter(L, R, l, r);
        if( cnt ) ans = int((ans + 1LL * cnt * fpow( int(b[i].first % y), x, y )) % y);
    }
    printf( "%d\n", ans );
}

void output_array() {
    for( int i = 0; i < sz; ++i ) {
        int t = b[i].second;
        while( t-- )
            dprint( "%lld ", b[i].first );
    }
    dprint("\n");
}

int main() {
    _w = scanf( "%d%d%d%d", &n, &m, &seed, &vmax );
    // dprint("Init:\n");
    for( int i = 1; i <= n; ++i ) {
        a[i] = rnd() % vmax + 1;
        // dprint("%lld ", a[i]);
    }
    // dprint("\n");
    prelude();
    for( int i = 1; i <= m; ++i ) {
        int op, l, r, x, y;
        op = rnd() % 4 + 1;
        l = rnd() % n + 1;
        r = rnd() % n + 1;
        if( l > r ) swap(l, r);
        if( op == 3 )
            x = rnd() % (r-l+1) + 1;
        else
            x = rnd() % vmax + 1;
        if( op == 4 )
            y = rnd() % vmax + 1;
        if( op == 1 ) solve1(l, r, x);
        else if( op == 2 ) solve2(l, r, x);
        else if( op == 3 ) solve3(l, r, x);
        else if( op == 4 ) solve4(l, r, x, y);
        // dprint("op = %d, l = %d, r = %d, x = %d, y = %d\n", op, l, r, x, y);
        // output_array();
    }
    return 0;
}
posted @ 2017-12-12 10:48 mlystdcall 阅读(...) 评论(...) 编辑 收藏