A. Pineapple Incident
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ted has a pineapple. This pineapple is able to bark like a bulldog! At time t (in seconds) it barks for the first time. Then every s seconds after it, it barks twice with 1 second interval. Thus it barks at times tt + st + s + 1, t + 2st + 2s + 1, etc.

Barney woke up in the morning and wants to eat the pineapple, but he can't eat it when it's barking. Barney plans to eat it at time x (in seconds), so he asked you to tell him if it's gonna bark at that time.

Input

The first and only line of input contains three integers ts and x (0 ≤ t, x ≤ 109, 2 ≤ s ≤ 109) — the time the pineapple barks for the first time, the pineapple barking interval, and the time Barney wants to eat the pineapple respectively.

Output

Print a single "YES" (without quotes) if the pineapple will bark at time x or a single "NO" (without quotes) otherwise in the only line of output.

Examples
input
3 10 4
output
NO
input
3 10 3
output
YES
input
3 8 51
output
YES
input
3 8 52
output
YES
Note

In the first and the second sample cases pineapple will bark at moments 3, 13, 14, ..., so it won't bark at the moment 4 and will bark at the moment 3.

In the third and fourth sample cases pineapple will bark at moments 3, 11, 12, 19, 20, 27, 28, 35, 36, 43, 44, 51, 52, 59, ..., so it will bark at both moments 51 and 52.

题意:一个序列 tt + st + s + 1, t + 2st + 2s + 1,。。。。。;判断x是否在里面;

思路:x-t对s取模,判断是否为0或1;特判(x-t)/s>0;

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define esp 0.00000000001
const int N=2e5+,M=1e6+,inf=1e9;
int main()
{
int x,y,z,i,t;
scanf("%d%d%d",&x,&y,&z);
if(x>z)
printf("NO\n");
else if(x==z)
printf("YES\n");
else if((z-x)%y==||(z-x)%y==)
{
if(z-x>=y)
printf("YES\n");
else
printf("NO\n");
}
else
printf("NO\n");
return ;
}
B. Barnicle
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Barney is standing in a bar and starring at a pretty girl. He wants to shoot her with his heart arrow but he needs to know the distance between him and the girl to make his shot accurate.

Barney asked the bar tender Carl about this distance value, but Carl was so busy talking to the customers so he wrote the distance value (it's a real number) on a napkin. The problem is that he wrote it in scientific notation. The scientific notation of some real number x is the notation of form AeB, where A is a real number and B is an integer and x = A × 10B is true. In our case A is between 0 and 9 and B is non-negative.

Barney doesn't know anything about scientific notation (as well as anything scientific at all). So he asked you to tell him the distance value in usual decimal representation with minimal number of digits after the decimal point (and no decimal point if it is an integer). See the output format for better understanding.

Input

The first and only line of input contains a single string of form a.deb where ad and b are integers and e is usual character 'e' (0 ≤ a ≤ 9, 0 ≤ d < 10100, 0 ≤ b ≤ 100) — the scientific notation of the desired distance value.

a and b contain no leading zeros and d contains no trailing zeros (but may be equal to 0). Also, b can not be non-zero if a is zero.

Output

Print the only real number x (the desired distance value) in the only line in its decimal notation.

Thus if x is an integer, print it's integer value without decimal part and decimal point and without leading zeroes.

Otherwise print x in a form of p.q such that p is an integer that have no leading zeroes (but may be equal to zero), and q is an integer that have no trailing zeroes (and may not be equal to zero).

Examples
input
8.549e2
output
854.9
input
8.549e3
output
8549
input
0.33e0
output
0.33

题意:将这个数改成整数或小数,。不能有前后导0;

思路:0.0e0这种数据小心点;

我yong码

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <sstream>
#include <vector>
#define PI acos(-1.0)
const int inf = (<<) - ;
using namespace std;
inline int get_int()
{
int r=;
char c;
while((c=getchar())!=' '&&c!='\n')
r=r*+c-'';
return r;
}
inline void out(int x)
{
if(x>)
{
out(x/);
}
putchar(x % + '');
putchar('\n');
}
/****************************************/ char a[],b[],c[],d[];
int xiao(int xx,int yy){return xx<yy?xx:yy;}
int main()
{
int aa,dd,bb,l,n,i,j,xx;
cin>>c;
n=strlen(c);
aa=c[]-'';
for(j=,i=;i<n;i++,j++){
if(c[i]=='e')
break;
d[j]=c[i];
}
bb=;
for(j=,i++;i<n;i++,j++)
bb=bb*+c[i]-'';
//aa=strlen(a);
//bb=strlen(b);
dd=strlen(d);
l=;
if(aa)
printf("%d",aa);
l=aa;
xx=xiao(bb,dd);
for(i=;i<xx;i++){
if(l==&&d[i]=='')
continue;
printf("%c",d[i]);
l=;
}
if(xx==dd){
for(i=dd;i<bb;i++)
printf("");
}
else{
if(l==)
printf("");
l=;
for(int i=xx;i<dd;i++){
if(d[i]>''){
l=;
break;
}
}
if(l){
printf(".");
for(i=xx;i<dd;i++)
printf("%c",d[i]);
} }
printf("\n");
}

代码

C. Lorenzo Von Matterhorn
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Barney lives in NYC. NYC has infinite number of intersections numbered with positive integers starting from 1. There exists a bidirectional road between intersections i and 2i and another road between i and 2i + 1 for every positive integer i. You can clearly see that there exists a unique shortest path between any two intersections.

Initially anyone can pass any road for free. But since SlapsGiving is ahead of us, there will q consecutive events happen soon. There are two types of events:

1. Government makes a new rule. A rule can be denoted by integers vu and w. As the result of this action, the passing fee of all roads on the shortest path from u to v increases by w dollars.

2. Barney starts moving from some intersection v and goes to intersection u where there's a girl he wants to cuddle (using his fake name Lorenzo Von Matterhorn). He always uses the shortest path (visiting minimum number of intersections or roads) between two intersections.

Government needs your calculations. For each time Barney goes to cuddle a girl, you need to tell the government how much money he should pay (sum of passing fee of all roads he passes).

Input

The first line of input contains a single integer q (1 ≤ q ≤ 1 000).

The next q lines contain the information about the events in chronological order. Each event is described in form 1 v u w if it's an event when government makes a new rule about increasing the passing fee of all roads on the shortest path from u to v by w dollars, or in form 2 v u if it's an event when Barnie goes to cuddle from the intersection v to the intersection u.

1 ≤ v, u ≤ 1018, v ≠ u, 1 ≤ w ≤ 109 states for every description line.

Output

For each event of second type print the sum of passing fee of all roads Barney passes in this event, in one line. Print the answers in chronological order of corresponding events.

Example
input
7
1 3 4 30
1 4 1 2
1 3 6 8
2 4 3
1 6 1 40
2 3 7
2 2 4
output
94
0
32
Note

In the example testcase:

Here are the intersections used:

  1. Intersections on the path are 3, 1, 2 and 4.
  2. Intersections on the path are 4, 2 and 1.
  3. Intersections on the path are only 3 and 6.
  4. Intersections on the path are 4, 2, 1 and 3. Passing fee of roads on the path are 32, 32 and 30 in order. So answer equals to32 + 32 + 30 = 94.
  5. Intersections on the path are 6, 3 and 1.
  6. Intersections on the path are 3 and 7. Passing fee of the road between them is 0.
  7. Intersections on the path are 2 and 4. Passing fee of the road between them is 32 (increased by 30 in the first event and by 2 in the second).

题意:给你一颗二叉树;q个询问;

   1:在u->v经过的线段+权值;

  2:询问u->v的总权值;

思路:利用map标记;有点类似求lca的过程;

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define esp 0.00000000001
const int N=2e5+,M=1e6+,inf=1e9;
map<ll,ll>m;
ll check(ll x)
{
ll sum=;
while(x)
{
sum++;
x>>=;
}
return sum;
}
void update(ll u,ll v,ll w)
{
while(check(u)>check(v))
{
m[u]+=w;
u/=;
}
while(check(u)<check(v))
{
m[v]+=w;
v/=;
}
while(check(u)!=&&u!=v)
{
m[v]+=w;
m[u]+=w;
v>>=;
u>>=;
}
}
ll query(ll u,ll v)
{
ll ans=;
while(check(u)>check(v))
{
ans+=m[u];
u/=;
}
while(check(u)<check(v))
{
ans+=m[v];
v/=;
}
while(check(u)!=&&u!=v)
{
ans+=m[v];
ans+=m[u];
v>>=;
u>>=;
}
return ans;
}
int main()
{
ll x,y,z,i,t;
scanf("%I64d",&x);
for(i=;i<x;i++)
{
ll flag,u,v;
scanf("%I64d%I64d%I64d",&flag,&u,&v);
if(flag==)
{
ll w;
scanf("%I64d",&w);
update(u,v,w);
}
else
{
printf("%I64d\n",query(u,v));
}
}
return ;
}

Codeforces Round #362 (Div. 2) A.B.C的更多相关文章

  1. Codeforces Round #362 (Div. 2) C. Lorenzo Von Matterhorn (类似LCA)

    题目链接:http://codeforces.com/problemset/problem/697/D 给你一个有规则的二叉树,大概有1e18个点. 有两种操作:1操作是将u到v上的路径加上w,2操作 ...

  2. #map+LCA# Codeforces Round #362 (Div. 2)-C. Lorenzo Von Matterhorn

    2018-03-16 http://codeforces.com/problemset/problem/697/C C. Lorenzo Von Matterhorn time limit per t ...

  3. 【转载】【树形DP】【数学期望】Codeforces Round #362 (Div. 2) D.Puzzles

    期望计算的套路: 1.定义:算出所有测试值的和,除以测试次数. 2.定义:算出所有值出现的概率与其乘积之和. 3.用前一步的期望,加上两者的期望距离,递推出来. 题意: 一个树,dfs遍历子树的顺序是 ...

  4. Codeforces Round #362 (Div. 2)->B. Barnicle

    B. Barnicle time limit per test 1 second memory limit per test 256 megabytes input standard input ou ...

  5. Codeforces Round #362 (Div. 2)->A. Pineapple Incident

    A. Pineapple Incident time limit per test 1 second memory limit per test 256 megabytes input standar ...

  6. Codeforces Round #362 (Div. 2) B 模拟

    B. Barnicle time limit per test 1 second memory limit per test 256 megabytes input standard input ou ...

  7. Codeforces Round #362 (Div. 2) A 水也挂

    A. Pineapple Incident time limit per test 1 second memory limit per test 256 megabytes input standar ...

  8. Codeforces Round #362 (Div. 2)

    闲来无事一套CF啊,我觉得这几个题还是有套路的,但是很明显,这个题并不难 A. Pineapple Incident time limit per test 1 second memory limit ...

  9. Codeforces Round #362 (Div. 2) D. Puzzles

    D. Puzzles time limit per test 1 second memory limit per test 256 megabytes input standard input out ...

随机推荐

  1. cf 61E. Enemy is weak 树状数组求逆序数(WA) 分类: Brush Mode 2014-10-19 15:16 104人阅读 评论(0) 收藏

    #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> ...

  2. [解决方案] 当 IDENTITY_INSERT 设置为 OFF 时

    当 IDENTITY_INSERT 设置为 OFF 时,不能为表 'tbUser' 中的标识列插入显式值. 解决:这个情况是你的表里面,有一列数据类型是IDENTITY的,也就是数据库自动递增列对于自 ...

  3. JavaScript之工厂方式 构造函数方式 原型方式讲解

    一.工厂方式可以为一个对象,创建多个实例. var oCar = new Object; oCar.color = "red"; oCar.doors=4; oCar.mpg=23 ...

  4. 如何让360、遨游、猎豹等双核浏览器默认以webkit内核渲染网页?

    众知目前国内不少浏览器都自称双核,一般是 IE(Trident)+Webkit.因为 webkit 急速的体验和对 HTML5 的支持,有些情况下开发者可能希望用户优先甚至只使用 webkit 内核渲 ...

  5. poj 2187

    求凸包后枚举凸包上的点 #include <cstdio> #include <cstdlib> #include <cmath> #include <map ...

  6. logback日志配置文件代码示例

    <?xml version="1.0" encoding="UTF-8"?> <configuration scan="true&q ...

  7. LCT模板

    之前一直用的LCT模板,因为其实个人对LCT和Splay不是很熟,所以用起来总觉得略略的坑爹,过了一段时间就忘了,但事实上很多裸的LCT要改的东西是不多的,所以今天写了些注释,以后可能套起模板来会得心 ...

  8. (转)約瑟夫問題的兩個O(log n)解法

    約瑟夫問題的兩個O(log n)解法 這個是學習編程時的一個耳熟能詳的問題了: n個人(編號爲0,1,...,n-1)圍成一個圈子,從0號開始依次報數,每數到第m個人,這個人就得自殺, 之後從下個人開 ...

  9. CXF+Spring 搭建的WebService

    1.创建类 2.接口编写 package com.fan; import javax.jws.WebService; @WebService public interface IHelloWorld ...

  10. java 如何连接MySql数据库

    利用jdbc方式连接数据库. 1.添加mysql驱动jar包 我用的是这个驱动包mysql-connector-java-5.1.26-bin.jar 添加方式: 2.加载MySql驱动类 priva ...