BNUZ寒假训练1.29

A题：
You are given an integer n. Check if n has an odd divisor, greater than one (does there exist such a number x (x>1) that n is divisible by x and x is odd).

For example, if n=6, then there is x=3. If n=4, then such a number does not exist.
解题思路：一个数字如果是2的n次方的话就不能被奇数整除，反之则能被奇数整除。先判断一个数是不是偶数，如果是的话将一个数不断的除2，能被2除尽，那么就打印NO。反之则打印YES
如8,8%20,8/=2,4/2=2,2/=2,最后是1，打印NO。10,10%20,10/=2,5%2!=0. 5!=1，打印YES.

#include 
#define ll long long 
int main()
{
     
	ll n,t;
	while(~scanf("%lld",&t))
	{
     
	while(t--){
     
	scanf("%lld",&n);
	while(n%2==0)n/=2;
	if(n==1)printf("NO\n");
	else printf("YES\n");				
	}
	}
	return 0;
}

B题：
Polycarp remembered the 2020-th year, and he is happy with the arrival of the new 2021-th year. To remember such a wonderful moment, Polycarp wants to represent the number n as the sum of a certain number of 2020 and a certain number of 2021.

For example, if:

n=4041, then the number n can be represented as the sum 2020+2021;
n=4042, then the number n can be represented as the sum 2021+2021;
n=8081, then the number n can be represented as the sum 2020+2020+2020+2021;
n=8079, then the number n cannot be represented as the sum of the numbers 2020 and 2021.
Help Polycarp to find out whether the number n can be represented as the sum of a certain number of numbers 2020 and a certain number of numbers 2021.
解题思路：水题一道，直接上代码

#include 
int main()
{
     
	int t;
	scanf("%d",&t);
	while(t--){
     
	int a,b,n;
	scanf("%d",&n);
	a=n/2020;
	b=n%2020;
	if(a<b) printf("NO\n");	 
	else    printf("YES\n");		
	}
	return 0;
}

C题：
You have an array a1,a2,…,an. All ai are positive integers.

In one step you can choose three distinct indices i, j, and k (i≠j; i≠k; j≠k) and assign the sum of aj and ak to ai, i. e. make ai=aj+ak.

Can you make all ai lower or equal to d using the operation above any number of times (possibly, zero)?
解题思路：
分两种情况：1.全部数字都小于d。
2.最小两个数字小于d。
那么就要找出两个最小值min1，min2;
方法1：冒泡排序，但是有可能超限。
方法2：线性查找，如下：

#include 
int main()
{
     
	int t;
	while(~scanf("%d",&t))
	{
     
	while(t--){
     
	int min1,min2,k,imin1,i,n,d,a[300];
	scanf("%d%d",&n,&d);
	k=1;
	for(i=0;i<n;i++){
     
	scanf("%d",&a[i]);
	if(i==0){
     min1=a[0];imin1=0;}
	if(min1>a[i]){
     
		min1=a[i];imin1=i;
	}
	if(a[i]>d){
     
		k=0;
	}
	}
	if(k){
     
		printf("YES\n");continue;
	}
	if(imin1==0)min2=a[1];
	else min2=a[0];
	for(i=0;i<n;i++){
     
		if(i!=imin1){
     
			if(min2>a[i]){
     
				min2=a[i];
			}
		}
	}
	if(min1+min2<=d)printf("YES\n");
	else printf("NO\n");		
	}			
	}	
	return 0;
}

D题：网上有答案，答案如下：

#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 7;
const int INF = 0x3f3f3f3f;
char s[maxn],t[maxn];
int len1,len2;
bool check1(int x,int len,char *s) {
     
    if(len % x != 0) return false;
    for(int i = 0;i < len;i++) {
     
        if(s[i] != s[i % x]) return false;
    }
    return true;
}
bool check3(int x,int y) {
     
    if(x != y) return false;
    for(int i = 0;i < x;i++) {
     
        if(s[i] != t[i]) return false;
    }
    return true;
}
int gcd(int n,int m) {
     
    return m == 0 ? n : gcd(m,n % m);
}
int main() {
     
    int T;scanf("%d",&T);
    while(T--) {
     
        scanf("%s%s",s,t);
        len1 = strlen(s),len2 = strlen(t);
        int l1 = len1,l2 = len2;
        //找s字符串的最小循环节 
        for(int i = 1;i <= len1;i++) {
     
            if(check1(i,len1,s)) {
     
                l1 = i;break;
            }
        }
        //找t字符串的最小循环节 
        for(int i = 1;i <= len2;i++) {
     
            if(check1(i,len2,t)) {
     
                l2 = i;break;
            }
        }
        //如果这俩字符串的最小循环节不一样
		  //就打印-1 
        if(check3(l1,l2)==0) {
     
            printf("-1\n");
        }//如果一样就找出他们的最小公倍数 
		  	else {
     
            len1 /= l1;
            len2 /= l2;
            int lcm = len1 * len2 / gcd(len1,len2);
            for(int i = 0;i < lcm;i++) {
     
                for(int j = 0;j < l1;j++) {
     
                    printf("%c",s[j]);
                }
            }
            printf("\n");
        }
    }
    return 0;
}