小陈的开学第十五周代码

1.Atcoder

1.Strings with the Same Length

题意：

Given are strings s and t of length N each, both consisting of lowercase English letters.

Let us form a new string by alternating the characters of S and the characters of T, as follows: the first character of S, the first character of T, the second character of S, the second character of T, …, the N-th character of S, the N-th character of T. Print this new string.

输入：

Input is given from Standard Input in the following format:

N
S T

1≤N≤100
|S|=|T|=N
S and T are strings consisting of lowercase English letters.

输出：

Print the string formed.

样例输入1：

2
ip cc

样例输出1：

icpc

样例输入2：

8
hmhmnknk uuuuuuuu

样例输出2：

humuhumunukunuku

样例输入3：

5
aaaaa aaaaa

样例输出3：

aaaaaaaaaa

程序代码：

//就是字符交替出现
#include<bits/stdc++.h>
using namespace std;
int main(){
	int n;
	cin>>n;
	char a[100+5],b[100+5];
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	for(int i=0;i<n;i++){
		cin>>b[i];
	}
	int t1=0,t2=0;
	for(int i=0;i<n*2;i++){
		if(i%2==0){
			cout<<a[t1];
			t1++;
		}else{
			cout<<b[t2];
			t2++;
		}	
	}
	return 0;
}

2.Snack

题意：

Takahashi is organizing a party.

At the party, each guest will receive one or more snack pieces.

Takahashi predicts that the number of guests at this party will be A or B.

Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted.

We assume that a piece cannot be divided and distributed to multiple guests.

输入：

Input is given from Standard Input in the following format:

A B

1≤A,B≤10^5
A≠B
All values in input are integers.

输出：

Print the minimum number of pieces that can be evenly distributed to the guests in both of the cases with A guests and B guests.

样例输入1：

2 3

样例输出1：

6

When we have six snack pieces, each guest can take three pieces if we have two guests, and each guest can take two if we have three guests.

样例输入2：

123 456

样例输出2：

18696

样例输入3：

100000 99999

样例输出3：

9999900000

程序代码：

//就是求最小公约数
#include<bits/stdc++.h>
using namespace std;
long long gcd(int a,int b){
    if(b==0)
        return a;
    return 
        gcd(b,a%b);
}
int main(){
	int a,b;
	cin>>a>>b;
	printf("%lld\n",a/gcd(a,b)*b);
	return 0;
}

3.Brick Break

题意：

We have N bricks arranged in a row from left to right.

The i-th brick from the left (1≤i≤N) has an integer ai written on it.

Among them, you can break at most N−1 bricks of your choice.

Let us say there are K bricks remaining. Snuke will be satisfied if, for each integer i(1≤i≤K), the i-th of those brick from the left has the integer i written on it.

Find the minimum number of bricks you need to break to satisfy Snuke’s desire. If his desire is unsatisfiable, print -1 instead.

输入：

Input is given from Standard Input in the following format:

N
a1 a2 ... aN

All values in input are integers.
1≤N≤200000
1≤ai≤N

输出：

Print the minimum number of bricks that need to be broken to satisfy Snuke’s desire, or print -1 if his desire is unsatisfiable.

样例输入1：

3
2 1 2

样例输出1：

1

If we break the leftmost brick, the remaining bricks have integers 1 and 2 written on them from left to right, in which case Snuke will be satisfied.

样例输入2：

3
2 2 2

样例输出2：

-1

In this case, there is no way to break some of the bricks to satisfy Snuke’s desire.

样例输入3：

10
3 1 4 1 5 9 2 6 5 3

样例输出3：

7

样例输入4：

1
1

样例输出4：

0

There may be no need to break the bricks at all.

解题思路：

这题就是问要去掉几个数字让序列变成123…的序列。
我错了好几次。
用a数组存第一次出现的位置，flag记录这串序列到几，最后别忘了要加上还没加上的。

程序代码：

#include<bits/stdc++.h>
using namespace std;
int a[200000+5];
int main(){
	int n;
	cin>>n;
	int flag=1;
	for(int i=1;i<=n;i++){
		int t;
		scanf("%d",&t);
		if(t==flag){
			if(a[t]==0){
				a[t]=i;
			}
			flag++;
		}
	}
	if(a[1]==0){
		cout<<"-1"<<endl;
		return 0;
	}
	long long num=0;
	a[0]=0;
	for(int i=1;i<=flag-1;i++){
		num+=(a[i]-a[i-1]-1);
	}
	num+=(n-a[flag-1]);//加上最后还没加上的
	printf("%lld\n",num);
	return 0;
}