【除法分块】power oj 2886

大力出奇迹

$$\begin{gathered} \sum _{i=l}^{r}x\%(k\ast i)=\sum _{i=l}^r(x-(k\ast i)\ast \lfloor \frac{x}{k\ast i}\rfloor ) \\ =x\ast (r-l+1)-\sum _{i=l}^r(k\ast i)\ast \lfloor \frac{x/k}{i}\rfloor \end{gathered}$$

后边$\sum _{i=l}^r(k\ast i)\ast \lfloor \frac{x/k}{i}\rfloor$就是除法分块的部分。

$\lfloor \frac{x/k}{i}\rfloor$在一段连续$[i]$区间是相同的

所以我们枚举 $L$ 令：$t=\lfloor \frac{x/k}{L}\rfloor$

然后找到 $t$ 相同的区间右端点 $R$：$R=t==0?r:min(x/k/t,r)$

这一块的值就是：$t\ast k\ast (R-L+1)\ast (R+L)/2$：就是把相同的 t 和 k 提出来，后边是连续区间 i 相加，也就是差值为1的等差序列求和。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <set>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll ;

ll read()
{
    
    
    ll x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {
    
     if(c == '-') f = -f; c = getchar(); }
    while(c >= '0' && c <= '9') {
    
     x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

const int maxN = 20;
const int maxM = 1003;
const ll mod = 1e9 + 7;

ll fpow(ll x, ll y) {
    
    
    ll ans = 1, base = x;
    while(y) {
    
    
        if(y & 1)
            ans *= base, ans %= mod;
        base *= base, base %= mod;
        y >>= 1;
    }
    return ans;
}

ll L, R, x, k;

int main()
{
    
    
    int T; cin >> T;
    while(T -- ) {
    
    
        L = read(), R = read(), x = read(), k = read();
        ll ans = (R - L + 1) * x % mod; x /= k;
        for(ll l = L, r, t; l <= R; l = r + 1) {
    
    
            t = x / l;
            r = t ? min(x / t, R) : R;
            ans -= t * k * (r - l + 1) * (l + r) % mod * fpow(2, mod - 2) % mod;
            ans = (ans % mod + mod) % mod;
        }
        cout << ans % mod << endl;
    }
    return 0;
}

基础除法分块：
P2261 [CQOI2007]余数求和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <set>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll ;

ll read()
{
    
    
    ll x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {
    
     if(c == '-') f = -f; c = getchar(); }
    while(c >= '0' && c <= '9') {
    
     x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

const int maxN = 20;
const int maxM = 1003;
const ll mod = 1e9 + 7;
ll n, k;

int main()
{
    
    
    n = read(), k = read();
    ll ans = n * k;
    for(ll l = 1, r, t; l <= n; l = r + 1 ) {
    
    
        t = k / l;
        r = t ? min(k / t, n) : n;
        ans -= t * (r - l + 1) * (r + l) >> 1;
    }
    cout << ans << endl;
    return 0;
}