Loj #6280. 数列分块入门 4 传送门

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对每一块记录总和s[i]，(注意：这里的s[i]不包括p[i]的值，当然，也可以写成包括的形式)，区间加时将不完整块暴力，对于完整块(不包括两边界，两边界即使完整也是用暴力)加上p[i] * m + s[i]，再取mod就可以了。详情请见代码：

#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] ){//l、r在同一块中直接暴力
        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;//左边界(不一定不完整)每个数加c
    for ( int i = r; p[r] == p[i]; --i ) a[i] += c, s[p[i]] += c;//右边界(也不一定完整)每个数加c
    for ( int i = p[l] + 1; i < p[r]; ++i ) v[i] += c;//中间的完整块加c
}

LL query( int l, int r, LL c ){
    if ( p[l] == p[r] ){//l、r在同一块中直接暴力
        LL ans(0);
        for ( int i = l; i <= r; ++i ) ans += ( a[i] + v[p[i]] ) % c, ans %= c;
        return ans;
    }
    LL ans(0);
    for ( int i = l; p[i] == p[l]; ++i ) ans += ( a[i] + v[p[i]] ) % c, ans %= c;//左边界取和
    for ( int i = r; p[i] == p[r]; --i ) ans += ( a[i] + v[p[i]] ) % c, ans %= c;//右边界取和
    for ( int i = p[l] + 1; i < p[r]; ++i ) ans += ( s[i] + v[i] * m ) % c, ans %= c;//中间完整块加上v[i] * m(完整块的大小都为m)与区间和s[i]
    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];//分成√n块，每一块s[i]加上初始值
    for ( int i = 1; i <= n; ++i ){
        scanf( "%lld%lld%lld%lld", &opt, &l, &r, &c );
        if ( opt ) printf( "%lld\n", query( l, r, c + 1 ) );
        else Add( l, r, c );
    }
    return 0;
}