【CF1207F】Remainder Problem

Description

【CF1207F】Remainder Problem

给定一个序列，支持如下操作：

• 将$x$位置的值加上$y$
• 求出$sumlimits_{i mod x= y}{a_i}$的值

Solution

 类似于分块的思路

可以肯定，$O(n^2)$肯定是T飞了

我们假设以T为界，预处理出模数小于T的答案，查询时如果模数小于T就直接返回，时间复杂度为$O(T)$

如果模数大于T就暴力计算，时间复杂度为$O(N/T)$

显然，当$T=sqrt{N}$时这两个复杂度相等。

所以总的复杂度为$O(msqrt{N})$

Code

#include <bits/stdc++.h>
namespace shl {
    typedef long long ll;
    const int N = 5e5 + 10;
    int a[N], ans[800][800];
    int n, m;
    inline int read() {
        int ret = 0, op = 1;
        char c = getchar();
        while (!isdigit(c)) {
            if (c == '-') op = -1; 
            c = getchar();
        }
        while (isdigit(c)) {
            ret = (ret << 3) + (ret << 1) + c - '0';
            c = getchar();
        }
        return ret * op;
    }
    int main() {
        n = 500000;
        m = read();
        int size = sqrt(n);
        while (m--) {
            int op; int x, y;
            std :: cin >> op; x = read(), y = read();
            if (op == 2) {
                if (x <= size) printf("%d
", ans[x][y]);
                else {
                    int sum = 0;
                    for (register int i = y; i <= n; i += x) sum += a[i];
                    printf("%d
", sum);
                }
            }
            else {
                for (register int i = 1; i <= size; ++i) ans[i][x % i] += y;
                a[x] += y; 
            }    
        }
        return 0;
    }
}
int main() {
    shl :: main();
    return 0;
}