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loj6280

题意概括

区间加法,区间求和.

写在前面

这题…也与分块1如出一辙…

正题

当有修改时,对于完整的块,直接维护一个数组$v$记录整个块加过的数(每块共同的加数)与s记录每个块的和(不算共同加数),不完整的就直接暴力在原数组$a$上直接加,并且别忘了给$s$也加上.询问时,对于不完整的块,直接暴力加(别忘了共同加数),完整的块对于每个区间$ans$加上(区间和 + 共同加数$\times$大小)就可以了.

代码

#include<cstdio>
#include<cmath>
using namespace std;
#define MAXN 50005
#define LL long long

int n, t, m;
LL a[MAXN], p[MAXN], v[500], s[500];
LL opt, l, r, c;

void Add( int l, int r, int c ){
    if ( p[l] == p[r] ){
        for ( int i = l; i <= r; ++i ) a[i] += c, s[p[i]] += c;
        return;
    }
    for ( int i = l; p[l] == p[i]; ++i ) a[i] += c, s[p[i]] += c;
    for ( int i = r; p[r] == p[i]; --i ) a[i] += c, s[p[i]] += c;
    for ( int i = p[l] + 1; i < p[r]; ++i ) v[i] += c;
}

LL query( int l, int r, LL c ){
    if ( p[l] == p[r] ){
        LL ans(0);
        for ( int i = l; i <= r; ++i ) ans += ( a[i] + v[p[i]] ) % c, ans %= c;
        return ans;
    }
    int ans(0);
    for ( int i = l; p[i] == p[l]; ++i ) ans += ( a[i] + v[p[i]] ) % c, ans %= c;
    for ( int i = r; p[i] == p[r]; --i ) ans += ( a[i] + v[p[i]] ) % c, ans %= c;
    for ( int i = p[l] + 1; i < p[r]; ++i ) ans += ( s[i] + v[i] * m ) % c, ans %= c;
    return ans % c;
}

int main(){
    scanf( "%d", &n ); m = (int)sqrt(n);
    for ( int i = 1; i <= n; ++i )
        scanf( "%lld", &a[i] ), p[i] = ( i - 1 ) / m + 1, s[p[i]] += a[i];
    for ( int i = 1; i <= n; ++i ){
        scanf( "%d%lld%lld%lld", &opt, &l, &r, &c );
        if ( opt ) printf( "%lld\n", query( l, r, c + 1 ) );
        else Add( l, r, c );
    }
}