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loj6278

题意概括

区间加法,区间询问小于一个数的个数.

正题

对于每个块,除原数组之外用一个vector来有序地存储所有数.当区间加时,对于每个完整块维护共同加数,对于不完整的块直接暴力加上再重新排序.当询问时,对于每个完整块在vector中二分,对于不完整的,直接暴力计数.

代码

#include<cstdio>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
#define MAXN 50005

int n, m, a[MAXN], p[MAXN], b[500], mm;
vector<int> v[500];
int opt, l, r, c;

int EF( vector<int> vec, int x ){//自力更生,手打二分
    int l, r, mid, ans(-1);
    l = 0; r = vec.size() - 1;
    while( l <= r ){
        mid = ( l + r ) >> 1;
        if ( vec[mid] < x ){
            ans = mid;
            l = mid + 1;
        }
        else r = mid - 1;
    }
    return ans + 1;
}

int query( int l, int r, int c ){
    int ans(0);

    if ( p[l] == p[r] ){
        for ( int i = l; i <= r; ++i )
            if ( a[i] + b[p[l]] < c ) ans++;
        return ans;
    }

    for ( int i = l; p[i] == p[l]; ++i )
        if ( a[i] + b[p[i]] < c ) ans++;
    for ( int i = r; p[i] == p[r]; --i )
        if ( a[i] + b[p[i]] < c ) ans++;
    for ( int i = p[l] + 1; i <= p[r] - 1; ++i )
        ans += EF( v[i], c - b[i] );
    return ans;
}

void re( int x ){
    v[x].clear();
    int be(( x - 1 ) * m + 1);
    for ( int i = be; p[i] == p[be]; i++ ) v[x].push_back( a[i] );
    sort( v[x].begin(), v[x].end() );
}

void Add( int l, int r, int c ){
    if ( p[l] == p[r] ){
        for ( int i = l; i <= r; ++i ) a[i] += c;
        re( p[l] ); return; 
    }

    for ( int i = l; p[i] == p[l]; ++i ) a[i] += c;
    re(p[l]);//重排.实际上可以归并排序,或者要用时再临时排序,这里偷了懒QAQ
    for ( int i = r; p[i] == p[r]; --i ) a[i] += c;
    re(p[r]);
    for ( int i = p[l] + 1; i < p[r]; ++i ) b[i] += c;
}

int main(){
    scanf( "%d", &n ); m = (int)sqrt(n);

    for ( int i = 1; i <= n; ++i ) p[i] = ( i - 1 ) / m + 1, mm = p[i];
    for ( int i = 1; i <= n; ++i ) scanf( "%d", &a[i] );
    for ( int i = 1; i <= n; ++i ) v[p[i]].push_back(a[i]);
    for ( int i = 1; i <= mm; ++i ) sort( v[i].begin(), v[i].end() );

    for ( int i = 1; i <= n; ++i ){
        scanf( "%d%d%d%d", &opt, &l, &r, &c );
        if ( opt ) printf( "%d\n", query( l, r, c * c ) );
        else Add( l, r, c );
    }
    return 0;
}

总结

比起其他算法老长老长的代码,分块算法的灵活在本题中得到体现QAQ