[BZOJ-1968]约束研究

Description

求出1~n的约数个数和。

Solution

这是一道[AHOI2005]的题。
我们考虑每个约数出现了多少次,枚举约数即可。

Notice

没什么要注意的。

Code

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#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define Rep(i, a, b) for (int i = (a); i < (b); i++)
#define travel(i, u) for (int i = head[u]; ~i; i = edge[i].next)

const ll INF = 1e9, Mo = 998244353;
const int N = 2000000;
const double eps = 1e-6;
namespace slow_IO
{
ll read()
{
ll x = 0; int zf = 1; char ch = getchar();
while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return x * zf;
}
void write(ll y)
{
if (y < 0) putchar('-'), y = -y;
if (y > 9) write(y / 10);
putchar(y % 10 + '0');
}
}
using namespace slow_IO;

int sqz()
{
int n = read(), ans = 0;
rep(i, 1, n) ans += n / i;
printf("%d\n", ans);
return 0;
}
文章目录
  1. 1. Description
  2. 2. Solution
  3. 3. Notice
  4. 4. Code