出神入化的分塊 Educational Codeforces Round 71 (Rated for Div. 2) E Remainder Problem

Educational Codeforces Round 71 (Rated for Div. 2) E Remainder Problem

題意: 兩種操作，一種 a x + y a_x+y ax​+y,第二種查詢

∑ i = y 5 e 5 a i , ( i % x = = y ) \sum_{i=y}^{5e5}a_i,(i\%x==y) i=y∑5e5​ai​,(i%x==y)

所有模 x x x等于 y y y位置的和.

題解: % x = y \%x=y %x=y 這不就是分塊麼，就是分塊的性質啊，直接處理不就OK了？？，想什麼線段樹。真6.

#include "bits/stdc++.h"
 
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> P;
#define VNAME(value) (#value)
#define bug printf("*********\n");
#define debug(x) cout<<"["<<VNAME(x)<<" = "<<x<<"]"<<endl;
#define mid ((l + r) >> 1)
#define chl 2 * k + 1
#define chr 2 * k + 2
#define lson l, mid, chl
#define rson mid + 1, r, chr
#define eb(x) emplace_back(x)
#define pb(x) emplace_back(x)
#define mem(a, b) memset(a, b, sizeof(a));
 
const LL mod = (LL) 1e9 + 7;
const int maxn = (int) 1e3 + 5;
const LL INF = 0x7fffffff;
const LL inf = 0x3f3f3f3f;
const double eps = 1e-8;
 
#ifndef ONLINE_JUDGE
clock_t prostart = clock();
#endif
 
void f() {
#ifndef ONLINE_JUDGE
    freopen("../data.in", "r", stdin);
#endif
}
 
//typedef __int128 LLL;
 
template<typename T>
void read(T &w) {//讀入
    char c;
    while (!isdigit(c = getchar()));
    w = c & 15;
    while (isdigit(c = getchar()))
        w = w * 10 + (c & 15);
}
 
template<typename T>
void output(T x) {
    if (x < 0)
        putchar('-'), x = -x;
    int ss[55], sp = 0;
    do
        ss[++sp] = x % 10;
    while (x /= 10);
    while (sp)
        putchar(48 + ss[sp--]);
}
 
int n, m;
 
const int B = 1e3;
LL q[maxn][maxn];
LL a[maxn * 1000];
 
int main() {
    f();
    read(n);
    while (n--) {
        int op, x, y;
        scanf("%d%d%d", &op, &x, &y);
        if (op == 1) {
            a[x] += y;
            for (int i = 1; i <= B; i++) {
                q[i][x % i] += y;
            }
        } else {
            if (x <= B) {
                printf("%lld\n", q[x][y]);
            } else {
                LL res = 0;
                while (y <= 500000) {
                    res += a[y];
                    y += x;
                }
                printf("%lld\n", res);
            }
        }
    }
 
#ifndef ONLINE_JUDGE
    cout << "運作時間:" << 1.0 * (clock() - prostart) / CLOCKS_PER_SEC << endl;
#endif
    return 0;
}