Loj #6280. 數列分塊入門 4 傳送門
對每一塊記錄總和s[i],(注意:這裏的s[i]不包括p[i]的值,當然,也可以寫成包括的形式),區間加時將不完整塊暴力,對於完整塊(不包括兩邊界,兩邊界即使完整也是用暴力)加上p[i] * m + s[i],再取mod就可以了。詳情請見代碼:
#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在同一塊中直接暴力
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;//左邊界(不一定不完整)每個數加c
for ( int i = r; p[r] == p[i]; --i ) a[i] += c, s[p[i]] += c;//右邊界(也不一定完整)每個數加c
for ( int i = p[l] + 1; i < p[r]; ++i ) v[i] += c;//中間的完整塊加c
}
LL query( int l, int r, LL c ){
if ( p[l] == p[r] ){//l、r在同一塊中直接暴力
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;//左邊界取和
for ( int i = r; p[i] == p[r]; --i ) ans += ( a[i] + v[p[i]] ) % c, ans %= c;//右邊界取和
for ( int i = p[l] + 1; i < p[r]; ++i ) ans += ( s[i] + v[i] * m ) % c, ans %= c;//中間完整塊加上v[i] * m(完整塊的大小都爲m)與區間和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塊,每一塊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;
}