A. Slackline Adventure（思维 + 莫比乌斯）（2018-2019 ACM-ICPC Brazil Subregional Programming Contest）

A. Slackline Adventure

考虑枚举每个矩形的长跟宽，再统计这个矩形能在坐标轴上出现几次（同行相邻同列相邻的单独算），然后有如下式子：

2×∑i=1n−1∑j=1m−1(n−i)(m−j)[gcd⁡(i,j)=1][L2≤i2+j2≤R2]2 \times \sum_{i = 1} ^{n - 1} \sum_{j = 1} ^{m - 1}(n - i)(m - j)[\gcd(i, j) = 1][L ^ 2 \leq i ^ 2 + j ^ 2 \leq R ^ 2]\\ 2×i=1∑n−1j=1∑m−1(n−i)(m−j)[gcd(i,j)=1][L2≤i2+j2≤R2]
设F(n,m,T2)=∑i=1n−1∑j=1m−1(n−i)(m−j)[gcd⁡(i,j)=1,i2+j2≤T2]F(n, m, T ^ 2) = \sum\limits\limits_{i = 1} ^{n - 1} \sum\limits_{j = 1} ^{m - 1} (n - i)(m - j)[\gcd(i, j) = 1, i ^ 2 + j ^ 2 \leq T ^ 2]F(n,m,T2)=i=1∑n−1j=1∑m−1(n−i)(m−j)[gcd(i,j)=1,i2+j2≤T2]，则ans=F(n,m,R2)−F(n,m,L2−1)ans = F(n,m, R ^ 2) - F(n, m, L ^ 2 - 1)ans=F(n,m,R2)−F(n,m,L2−1)。

∑k=1n−1μ(k)∑i=1n−1k∑j=1m−1k(n−ik)(m−jk)[i2k2+j2k2≤T2]∑k=1n−1μ(k)∑i=1n−1k(n−ik)calc()\sum_{k = 1} ^{n - 1} \mu(k) \sum_{i = 1} ^{\frac{n - 1}{k}} \sum_{j = 1} ^{\frac{m - 1}{k}} (n - ik)(m - jk)[i ^ 2 k ^ 2 + j ^ 2 k ^ 2 \leq T ^ 2]\\ \sum_{k = 1} ^{n - 1} \mu(k) \sum_{i = 1} ^{\frac{n - 1}{k}} (n - ik) calc()\\ k=1∑n−1μ(k)i=1∑kn−1j=1∑km−1(n−ik)(m−jk)[i2k2+j2k2≤T2]k=1∑n−1μ(k)i=1∑kn−1(n−ik)calc()
calc()calc()calc()就是当已知i,k,Ti, k, Ti,k,T，有多少jjj满足i2k2+j2k2≤T2i ^ 2 k ^ 2 + j ^ 2 k ^ 2 \leq T ^ 2i2k2+j2k2≤T2，的m−jkm - jkm−jk求和，整体复杂度O(k×nlog⁡n)O(k \times n \log n)O(k×nlogn)，kkk是个比较小的常数

今天VP的时候碰到的，到了一直没看这题，到了最后才发现是个数论题。

说一点点坑点吧，注意函数不能写成f(n,m,T)=∑i=1n−1∑j=1m−1(n−i)(m−j)[gcd⁡(i,j)=1,i2+j2≤T2]f(n, m, T) = \sum\limits\limits_{i = 1} ^{n - 1} \sum\limits_{j = 1} ^{m - 1} (n - i)(m - j)[\gcd(i, j) = 1, i ^ 2 + j ^ 2 \leq T ^ 2]f(n,m,T)=i=1∑n−1j=1∑m−1(n−i)(m−j)[gcd(i,j)=1,i2+j2≤T2]，

然后答案是f(n,m,R)−f(n,m,L−1)f(n, m, R) - f(n, m, L - 1)f(n,m,R)−f(n,m,L−1)，这样会使满足条件得满足条件的一些数(L−1)2<i2k2+j2k2<L2(L - 1) ^ 2 <i ^ 2 k ^ 2 + j ^ 2 k ^ 2 < L ^ 2(L−1)2<i2k2+j2k2<L2，

没有被删去而算在贡献里了，会导致答案出错，就因为这个调了一晚上，一直不知道哪里错了，，，

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10, mod = 1e9 + 7;int prime[N], mu[N], cnt, n, m, L, R;bool st[N];inline int add(int x, int y) {return x + y < mod ? x + y : x + y - mod;
}inline int sub(int x, int y) {return x >= y ? x - y : x - y + mod;
}void init() {mu[1] = 1;for (int i = 2; i < N; i++) {if (!st[i]) {mu[i] = -1;prime[++cnt] = i;}for (int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if (i % prime[j] == 0) {break;}mu[i * prime[j]] = -mu[i];}}
}int calc(int i, int k, long long T) { // i * i * k * k + j * j * k * k <= Tint l = 0, r = (m - 1) / k;while (l < r) {int mid = l + r + 1 >> 1;if (1ll * i * i * k * k + 1ll * mid * mid * k * k > T) {r = mid - 1;}else {l = mid;}}return 1ll * (m - k + m - l * k) * l / 2 % mod;
}int f(long long T) {int ans = 0;for (int k = 1, lim, res; k <= n - 1; k++) {if (mu[k] == 0) {continue;}lim = (n - 1) / k, res = 0;for (int i = 1; i <= lim; i++) {res = add(res, 1ll * (n - i * k) * calc(i, k, T) % mod);}if (mu[k] == 1) {ans = add(ans, res);}else {ans = sub(ans, res);}}ans = 2ll * ans % mod;if (T >= 1) {ans = add(ans, add(1ll * n * (m - 1) % mod, 1ll * (n - 1) * m % mod));}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);init();scanf("%d %d %d %d", &n, &m, &L, &R);printf("%d\n", sub(f(1ll * R * R), f(1ll * L * L - 1)));return 0;
}

A. Slackline Adventure（思维 + 莫比乌斯）（2018-2019 ACM-ICPC Brazil Subregional Programming Contest）相关推荐

2019-2020 ACM ICPC Brazil Subregional Programming Contest E.Exhibition of Clownfish
我搬运我自己应该算原创吧题目题意题解题意某水族馆有一种神奇的
A - Multiplication Dilemma (思维)（ 2018 ACM ICPC Arabella Collegiate Programming Contest）
滴答滴答---题目链接 Multiplication operation is not always easy! For example, it is hard to calculate 27 × 2 ...
2018 ACM ICPC Arabella Collegiate Programming Contest A
Multiplication operation is not always easy! For example, it is hard to calculate 27 × 20 using your ...
2019 ACM - ICPC 上海网络赛 E. Counting Sequences II （指数型生成函数）
繁凡出品的全新系列:解题报告系列 -- 超高质量算法题单,配套我写的超高质量题解和代码,题目难度不一定按照题号排序,我会在每道题后面加上题目难度指数(1∼51 \sim 51∼5),以模板题难度 11 ...
2019 ACM - ICPC 西安邀请赛 B. Product （杜教筛）简单数论（bushi）
G.(2019 ACM/ICPC 全国邀请赛(西安)B) Product Weblink https://nanti.jisuanke.com/t/39269 Problem && S ...
2019 ACM - ICPC 全国邀请赛（西安）题解（9 / 13）
The 2019 ACM-ICPC China Shannxi Provincial Programming Contest 目录 The 2019 ACM-ICPC China Shannxi Pr ...
2019, XII Samara Regional Intercollegiate Programming Contest 解题报告
2019 XII Samara Regional Intercollegiate Programming Contest 传送门 A. Rooms and Passages 题意: 有 n+1n+1n ...
The 2019 ACM-ICPC China Shannxi Provincial Programming Contest题解
以下所有AC题解程序来自"仙客传奇"团队. A. Tasks AC的C++语言程序: /* JSK-39268 Tasks */#include <bits/stdc++.h ...
2019 ACM - ICPC 全国邀请赛（南昌）题解（9 / 12）
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划目录 The 2019 ICPC China Nanchang National Invitation ...

A. Slackline Adventure（思维 + 莫比乌斯）（2018-2019 ACM-ICPC Brazil Subregional Programming Contest）

A. Slackline Adventure

A. Slackline Adventure（思维 + 莫比乌斯）（2018-2019 ACM-ICPC Brazil Subregional Programming Contest）相关推荐

最新文章

热门文章