A .Banana Bananas are the favoured food of monkeys. In the forest, there is a Banana Company that provides bananas from different places. The company has two lists. The first list records the types of bananas preferred by different monkeys, and the s

Banana Bananas are the favoured food of monkeys. In the forest, there is a Banana Company that provides bananas from different places. The company has two lists. The first list records the types of bananas preferred by different monkeys, and the seco

2017 ACM-ICPC 亚洲区(乌鲁木齐赛区)网络赛  H Skiing In this winter holiday, Bob has a plan for skiing at the mountain resort. This ski resort has MM different ski paths and NN different flags situated at those turning points. The ii-th path from the S_iS​i​​-th f

「ICPC2017 WF」Money for Nothing 传送门 我们可将生产商和消费商都看成二维平面上的点,其坐标分别为 \((d_i,p_i)\),\((e_i,q_i)\). 那么问题转变为: 给定平面上的 \(m\) 个 \(A\) 类点 \((d_i,p_i)\),以及 \(n\) 个 \(B\) 类点 \((e_i,q_i)\).求选择一个 \(A\) 类点作为矩形左下角,一个 \(B\) 类点作为矩形右上角所能得到的最大面积.若不存在这样的矩形就输出零. 考虑如下几种情况. 在

H.Skiing In this winter holiday, Bob has a plan for skiing at the mountain resort. This ski resort has MM different ski paths and NN different flags situated at those turning points. The ii-th path from the S_iS​i​​-th flag to the T_iT​i​​-th flag ha

这个题B站上面有这题很完整的分析和证明,你实在不懂,可以看看这个视频  https://www.bilibili.com/video/av19849697?share_medium=android&share_source=qq&bbid=00561DA9-126A-4190-88A8-2B9DD5DAFEB512211infoc&ts=1533979978895    视频里面讲的很清楚,我再重复一下把,就是说,给N个圆上的点,在这个圆内部会产生多少个点呢?,很简单C(N,4);

题目描述: Life is a journey, and the road we travel has twists and turns, which sometimes lead us to unexpected places and unexpected people. Now our journey of Dalian ends. To be carefully considered are the following questions. Next month in Xian, an e

2017-09-09 16:41:28 writer:pprp 题意很好理解就不说了,实现比较清晰,选择邻接表来做 但是我用的是链表来实现的,所以导致出现了很多问题,最后卡的最长时间的一个问题是 应该从1开始而不是从0开始,读题应该自习一点: 题目如下: Bananas are the favoured food of monkeys. In the forest, there is a Banana Company that provides bananas from different pl

题目描述 Consider a triangle of integers, denoted by T. The value at (r, c) is denoted by Tr,c , where 1 ≤ r and 1 ≤ c ≤ r. If the greatest common divisor of r and c is exactly 1, Tr,c = c, or 0 otherwise.Now, we have another triangle of integers, denote

题目链接:https://nanti.jisuanke.com/t/16957 题目: In this winter holiday, Bob has a plan for skiing at the mountain resort. This ski resort has MMM different ski paths and NNN different flags situated at those turning points. The iii-th path from the SiS_i

#include<bits/stdc++.h>using namespace std;const long long mod = 998244353;typedef const long long ll;vector<long long>p;long long inv(long long x,long long y)//快速幂求逆元模板(以乘代除){    long long r=1;    while(y>0)    {        if(y&1)        

#include<bits/stdc++.h>using namespace std;int f[200010];//代表元long long rl[200010];//记rl[i]为结点i到根的路径上所有边的亦或和(这里可以忽视 x(0)的情况)void init(int n){    for(int i=1;i<=n;i++)        f[i]=i,rl[i]=0;//初始化}int fd(int x)//并查集模板{    if(f[x]==x)        return

#include<bits/stdc++.h>using namespace std;double c[110][110];double g[110];double dp[110][110];int n,m,k;void cnm(){    c[0][0]=1;    for(int i=1;i<=100;i++)    {        c[i][0]=1;        c[i][i]=1;        for(int j=1;j<i;j++)        {       

题目连接 : https://nanti.jisuanke.com/t/A1256 Life is a journey, and the road we travel has twists and turns, which sometimes lead us to unexpected places and unexpected people. Now our journey of Dalian ends. To be carefully considered are the following

额,只是一道签到题,emmm直接代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ; int c[maxn]; int d[maxn]; int main(){ int t; scanf("%d", &t); while (t--){ int n, m; scanf("%d %d", &n, &

题目大意是有一堆猴子,然后每个猴子都有自己喜欢的香蕉类型,然后香蕉会在指定的位置,问每个猴子能不能在每个地方吃到自己喜欢的香蕉. 其实直接暴力即可(因为最大最大最大是50) 上代码: #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; ][]; ][]; ][]; int n, m; int main() { int T; scanf("%d",

题目 https://nanti.jisuanke.com/t/16958 题意 给定一个n(n<=500)个点的无向图,给每条边黑白染色,输出同色三角形最少的个数和对应的方案 分析 首先考虑给定一个染色完毕的无向图,如何求同色三角形的个数 同色三角形的个数=总的个数-异色三角形的个数 而一个异色三角形对应两个异色角,所以我们可以通过算异色角的个数来计算异色三角形的个数 而异色角是有一个固定的点i引出去的n-1条边所决定的 设某个点i有$x_i$条1边,有$n-1-x_i$条2边 可以发现异色角

Life is a journey, and the road we travel has twists and turns, which sometimes lead us to unexpected places and unexpected people. Now our journey of Dalian ends. To be carefully considered are the following questions. Next month in Xian, an essenti

[链接]h在这里写链接 [题意] 让你维护字符串的一段区间内T子串的个数. [题解] 因为t不大,所以. 暴力维护一下a[i]就好. a[i]表示的是S串从i位置开始,能和T串匹配几个字符. 用树状数组维护区间内a[i]==lent的个数就好. 修改,还是暴力改就行.只会影响到那几个位置的. [错的次数] 0 [反思] 匹配的部分写在一个函数里面比较好,用起来比较方便. [代码] #include <bits/stdc++.h> using namespace std; const int N

[链接]h在这里写链接 [题意] 在这里写题意 [题解] 莫队算法+树状数组. 区间增加1或减少1. 对逆序对的影响是固定的. (用冒泡排序变成升序的交换次数,就是逆序对的个数) [错的次数] 0 [反思] 在这了写反思 [代码] #include <bits/stdc++.h> using namespace std; const int N = 3e4; int n,m, a[N + 10], l, r; long long ans[N + 10],now = 0; struct BI {