CF Round#240题解
第一次参加CF的比赛，MSK19.30，四个小时的时差真心累，第一次CODE到这么夜……
一开始做了A，C两题，后来做B题的时候我体力和精神集中度就很低了，导致一直WA在4……
今天起床后再刷B，终于过了……坑爹。
来看题目：
Mashmokh works in a factory. At the end of each day he must turn off all of the lights.
The lights on the factory are indexed from 1 to n. There are n buttons in Mashmokh's room indexed from 1 to n as well. If Mashmokh pushes button with index i, then each light with index not less than i that is still turned on turns off.
Mashmokh is not very clever. So instead of pushing the first button he pushes some of the buttons randomly each night. He pushed m distinct buttons b1, b2, ..., bm (the buttons were pushed consecutively in the given order) this night. Now he wants to know for each light the index of the button that turned this light off. Please note that the index of button bi is actually bi, not i.
Please, help Mashmokh, print these indices.
The first line of the input contains two space-separated integers n and m (1 ≤ n, m ≤ 100), the number of the factory lights and the pushed buttons respectively. The next line contains m distinct space-separated integers b1, b2, ..., bm (1 ≤ bi ≤ n).
It is guaranteed that all lights will be turned off after pushing all buttons.
Output n space-separated integers where the i-th number is index of the button that turns the i-th light off.
5 4
4 3 1 2
1 1 3 4 4
5 5
5 4 3 2 1
1 2 3 4 5
In the first sample, after pressing button number 4, lights 4 and 5 are turned off and lights 1, 2 and 3 are still on. Then after pressing button number 3, light number 3 is turned off as well. Pressing button number 1 turns off lights number 1 and 2 as well so pressing button number 2 in the end has no effect. Thus button number 4 turned lights 4 and 5 off, button number 3 turned light 3 off and button number 1 turned light 1 and 2 off.
英文不好的我也看懂题目了……其实意思很简单，就是关灯，关编号i的灯大于等于i的灯就会灭掉，之后给出n,k,n为灯树木，k为操作次数，接下来k个是关的灯。
贪心轻松搞掂。memset(v,0,sizeof(v)),之后关i时候就将未标记的（v[m]==0）标记为v[m]=i(m>=i),水到不行。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int v[103];
int main(){
int n;
int m,i,j,tmp;
while(cin>>n>>m){
memset(v,0,sizeof(v));
for(i=0;i<m;i++){
cin>>tmp;
for(j=tmp;j<=n;j++){
if(v[j]==0) v[j]=tmp;
}
}
cout<<v[1];
for(i=2;i<=n;i++)
cout<<" "<<v[i];
cout<<endl;
}
return 0;
}
B题
Bimokh is Mashmokh's boss. For the following n days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back w tokens then he'll get 
dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has n numbers x1, x2, ..., xn. Number xi is the number of tokens given to each worker on the i-th day. Help him calculate for each of n days the number of tokens he can save.
The first line of input contains three space-separated integers n, a, b (1 ≤ n ≤ 105; 1 ≤ a, b ≤ 109). The second line of input contains n space-separated integers x1, x2, ..., xn (1 ≤ xi ≤ 109).
Output n space-separated integers. The i-th of them is the number of tokens Mashmokh can save on the i-th day.
5 1 4
12 6 11 9 1
0 2 3 1 1
3 1 2
1 2 3
1 0 1
1 1 1
1
0
我一开始以为简单地取余数（弱爆了），之后我写了几组数据
1 100 99
98
1 98 99
100
1 11 4
2
1 11 4
3
发现全错掉了…
其实题目意思就是求
求最小的w满足floor(a*w/b)为最大，题中的表述就是爱代币，更爱金钱。
我的思路就是：
如果a比b大，那么直接全部换掉，输出0。
否则先计算a*w/b的向下取整m,再计算m*b/a的向上取整p，输出w-p（为什么？不明白的朋友可以用上面的数据手写一下）
…………轻松ac,为什么当时我就意识迷糊到这种题目都过不去，懊恼。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
using namespace std;
int main(){
long long n,a,b;
long long i,j;
long m;
while(cin>>n>>a>>b){
for(i=0;i<n;i++){
cin>>j;
if(a>=b){
cout<<0<<" ";
}else{
m=(a*j)/b;
m=ceil( 1.0*(m*b)/a );
cout<<j-m<<" ";
}
}
cout<<endl;
}
return 0;
}
It's holiday. Mashmokh and his boss, Bimokh, are playing a game invented by Mashmokh.
In this game Mashmokh writes sequence of n distinct integers on the board. Then Bimokh makes several (possibly zero) moves. On the first move he removes the first and the second integer from from the board, on the second move he removes the first and the second integer of the remaining sequence from the board, and so on. Bimokh stops when the board contains less than two numbers. When Bimokh removes numbers x and y from the board, he gets gcd(x, y) points. At the beginning of the game Bimokh has zero points.
Mashmokh wants to win in the game. For this reason he wants his boss to get exactly k points in total. But the guy doesn't know how choose the initial sequence in the right way.
Please, help him. Find n distinct integers a1, a2, ..., an such that his boss will score exactly k points. Also Mashmokh can't memorize too huge numbers. Therefore each of these integers must be at most 109.
The first line of input contains two space-separated integers n, k (1 ≤ n ≤ 105; 0 ≤ k ≤ 108).
If such sequence doesn't exist output -1 otherwise output n distinct space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).
5 2
1 2 3 4 5
5 3
2 4 3 7 1
7 2
-1
gcd(x, y) is greatest common divisor of x and y.
老实说，我觉得C题是一题有意思的水题，并没有上面两题那么水，主要是你的思考方式。
题目是的意思就是 要你构造n长的互异序列 {Ai}.
使得序列 中，sum[ gcd(Ai,A(i+1) ) ] =k,i=1~ n-(n%1) 且i%1==1
就是前面开始每两个的gcd加起来为k（直到没有办法组成两个，即奇数时候剩下An时候）。
如果不可能的话输出-1。
首先来分析下，什么叫做不可能
很简单，因为gcd(a,b)最小也为1,当Pair数目p= (n-(n%1))/2大于k时候，那么肯定是-1.
这是下界不满足。
上界不满足，最大的gcd(a,b)会不会比k小呢？仔细看范围，只有当n=1时候，k>0才有可能，因为Ai可以达到10^9次方，而k最多也就10^8,n最都也就是10^5。
但是n=1,k=0是输出1（或者其他，随便……）n=1,k>0,输出-1。
-1的情况考虑完后，怎么构造这个序列呢？其实很简单，真的很简单。
我们Pair的数目p，p-1对互素，使得tmp=p-1,剩下用一对来构造A1,A2,gcd(A1,A2)=k-tmp即可。
我的构造方法是: A1=k-tmp,A2=2*A1,Ai=Ai-1+1.
因为 P 与 P+1总是互素(P>1,而A1，A2的构造刚好排除了P=1情况),轻松AC。
PS：不能忘记剪枝N=1,的情况，不然会WA 4
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main(){
int n,k;
cin>>n>>k;
int pair=n/2 ;
if(pair>k ){
cout<<-1<<endl; //下界
}else if(n==1){
if(k==0) {
cout<<1<<endl;
}else cout<<-1<<endl;
}else{
if(pair==k){
for(int i=1;i<=n;i++)
cout<<i<<" ";
cout<<endl;
}else{
pair--; //用一对来构造
int tmp=k-pair;
cout<<tmp<<" "<<2*tmp;
tmp=2*(k-pair);
for(int i=1;i<=n-2;i++){
cout<<" "<<tmp+i;
}
cout<<endl;
}
}
return 0;
}
待续。