2 柠檬咕咕咕

尚未进行身份认证

我要认证

暂无相关简介

等级
TA的排名 2w+

2020牛客暑期多校训练营(第八场)G Game SET

string tmp[5][5]{{"one","two","three"},{"diamond","squiggle","oval"},{"solid","striped","open"},{"red","green","purple"}};map<string,int>id;char s[100];int n,a[300][10];void solve(){ for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j+..

2020-08-04 21:37:38

2020牛客暑期多校训练营(第八场)I Interesting Computer Game

ll n;ll a[100005],b[100005];ll suma[100005];pair<__int128,__int128>stb[100005],sta[100005];__int128 ans;__int128 kmax;__int128 k;int main() { ll t=read(); for(ll iii=1; iii<=t; ++iii) { cin>>n; //lirun .

2020-08-04 21:19:53

2020/8/1日志

训练一个月,牛客打了七场,从一开始签到到现在开始切中难题,平均每场能出一两道题,cf现在是蓝名,hhh压力还是有的信息学一本通跟进阶分别在石室中学网站和牛客上刷了一半,最近补完cf跟牛客多校继续刷题,顺便复习一下数据结构的图论,然后继续牛客上的进阶...

2020-08-02 21:14:28

Educational Codeforces Round 92 (Rated for Div. 2) B. Array Walk(dp)

B. Array Walktime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou are given an array a1,a2,…,an, consisting of n positive integers.Initially you are standing at index 1 and have a score equal to a1

2020-08-02 21:11:22

Educational Codeforces Round 92 (Rated for Div. 2) A. LCM Problem(构造)

A. LCM Problemtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputLet LCM(x,y) be the minimum positive integer that is divisible by both x and y. For example, LCM(13,37)=481, LCM(9,6)=18.You are given

2020-08-02 20:42:22

2020牛客暑期多校训练营(第七场)H Dividing

void cal(ll n, ll k){ for (ll l = 2, r; l <= k; l = r + 1) { r = n / l; r = min(r, n); if (r) r = n / r; else r = k; r = min(r, k); int len = (r - l + 1) % mod; i..

2020-08-01 21:25:10

2020牛客暑期多校训练营(第七场)D Fake News

int main(){ int t=read(); while (t--) { cin>>n; if (n == 1||n==24) puts("Fake news!"); else puts("Nobody knows it better than me!"); } return 0;}

2020-08-01 21:18:35

2020牛客暑期多校训练营(第七场)B Mask Allocation

int t;int n,m; vector <int >ve;void sol(int a,int b) { if(a<b) swap(a,b); if(!b) return ;//1 for(int i=1; i<=b; i++) { ve.push_back(b); } sol(b,a-b);} int main() { t=read(); while(t--) { ve...

2020-08-01 21:16:24

CodeForces1388D Captain Flint and Treasure

D. Captain Flint and Treasuretime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputCaptain Fint is involved in another treasure hunt, but have found only one strange problem. The problem may be connected

2020-08-01 00:01:10

牛客多校第六场 C Combination of Physics and Maths

int main() { int t; cin>>t; while(t--) { ll n,m; double num=0,arr[205][205]; cin>>n>>m; for(ll i=0; i<n; i++) { for(ll j=0; j<m; j++) { cin>>arr[i][j];..

2020-07-28 16:35:21

牛客多校第六场 E-Easy Construction (构造)

题意:给出n,k,构造出一个n的全排列P,使得对于 1~n 中任意的数 i,P 都存在一个长为 i 的子区间,其和模 n 余 k。有解输出任意一组,无解输出 -1转载:https://blog.csdn.net/zjllll123/article/details/107606334首先如果要有解,k 必须是 n(n+1)/2 % n,因为长度为 n 的子区间只有一个,也就是 P 本身,而 P 本身的元素和就是 n(n+1)/2。然后假设 k 满足上述条件,此时是一定有解的。且k的值是固定的。k.

2020-07-28 15:11:42

Codeforces Round #659 (Div. 2) C. String Transformation 1

C. String Transformation 1time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputNote that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this v

2020-07-27 11:25:33

Codeforces Round #659 (Div. 2) A. Common Prefixes

A. Common Prefixestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe length of the longest common prefix of two strings s=s1s2…sn and t=t1t2…tm is defined as the maximum integer k (0≤k≤min(n,m)) su

2020-07-27 11:12:35

Codeforces Round #658 (Div. 2) C2. Prefix Flip (Hard Version)

C2. Prefix Flip (Hard Version)time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThis is the hard version of the problem. The difference between the versions is the constraint on n and the required num

2020-07-23 18:45:48

2020牛客暑期多校训练营(第四场)F Finding the Order

题目描述ZYB has a so-called smart brain: He can always point out the keypoint in a complex problem.There are two parallel lines AB and CD in a plane. {A,B,C,D}A,B,C,D are all distinct points.You only know the Euclidean Distances between {AC,AD,BC,BD}AC,AD,B

2020-07-21 19:46:26

2019HDU多校第二场 K - Keen On Everything But Triangle(主席树)

N sticks are arranged in a row, and their lengths are a1,a2,…,aN.There are Q querys. For i-th of them, you can only use sticks between li-th to ri-th. Please output the maximum circumference of all the triangles that you can make with these sticks, or pri

2020-07-17 16:26:41

2020牛客暑期多校训练营(第一场)F Infinite String Comparision

int main() { while(cin>>str1&&cin>>str2) { ll len1=strlen(str1); ll len2=strlen(str2); ll len=2*max(len1,len2); bool flag=0; for(ll i=0; i<len; i++) { if(str1[i%len1]==str2[i...

2020-07-16 21:38:58

2020牛客暑期多校训练营(第一场)J Easy Integration

void factor (){ ll i=1; fac[0]=1; for (;i<=2000000;i++) { fac[i]=fac[i-1]*i%MOD; }} ll power(ll a,ll n){ ll res=1; while (n) { if (n&1) { res=res*a%MOD; } a*=a; a%=MO..

2020-07-16 21:37:35

2020牛客暑期多校训练营(第二场)C Cover the Tree

题目描述Given an unrooted tree, you should choose the minimum number of chains that all edges in the tree are covered by at least one chain. Print the minimum number and one solution. If there are multiple solutions, print any of them.输入描述:The first line co

2020-07-15 21:57:59

2020牛客暑期多校训练营(第二场)F Fake Maxpooling(二维滑动窗口)

时间限制:C/C++ 3秒,其他语言6秒空间限制:C/C++ 262144K,其他语言524288K64bit IO Format: %lld题目描述Given a matrix of size n\times mn×m and an integer {k}k, where A_{i,j} = lcm(i, j)A i,j =lcm(i,j), the least common multiple of {i}i and {j}j. You should determine the sum of th

2020-07-13 21:51:43

查看更多

勋章 我的勋章
  • 领英
    领英
    绑定领英第三方账户获取
  • GitHub
    GitHub
    绑定GitHub第三方账户获取
  • 签到新秀
    签到新秀
    累计签到获取,不积跬步,无以至千里,继续坚持!
  • 技术圈认证
    技术圈认证
    用户完成年度认证,即可获得
  • 阅读者勋章Lv2
    阅读者勋章Lv2
    授予在CSDN APP累计阅读博文达到7天的你,是你的坚持与努力,使你超越了昨天的自己。
  • 持之以恒
    持之以恒
    授予每个自然月内发布4篇或4篇以上原创或翻译IT博文的用户。不积跬步无以至千里,不积小流无以成江海,程序人生的精彩需要坚持不懈地积累!
  • 勤写标兵Lv4
    勤写标兵Lv4
    授予每个自然周发布9篇以上(包括9篇)原创IT博文的用户。本勋章将于次周周三上午根据用户上周的博文发布情况由系统自动颁发。
  • 分享学徒
    分享学徒
    成功上传1个资源即可获取