A. Maximum GCD
题意：给你一个n 从1~n中找两个数 a、b 使得 gcd(a,b)最大
做法：答案：n/2
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=2e3+10;
int a[N],n;
int main()
{
int _=read();while(_--)
{
int n=read();
printf("%d\n",n/2);
}
}
B. GCD Compression
题意：t组数据 每组2n(n<=1000)个数a[i]，每次从a数组选两个数求和，加入到b数组中，共（n-1）个数，使得 n-1 个数的 gcd 大于1
做法：选取构造两数之和为偶数即可。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=2e3+10;
int a[N],n;
int main()
{
int _=read();while(_--)
{
queue<int>odd,ev;
n=read();
rep(i,1,2*n)
{
a[i]=read();
if(a[i]%2) odd.push(i);
else ev.push(i);
}
vector<pair<int,int> >ans;
while(ans.size()<n-1){
//printf("an:%d\n",ans.size());
if(odd.size()>=2) {
int x=odd.front();odd.pop();
int y=odd.front();odd.pop();
ans.push_back({x,y});
}
if(ans.size()==n-1) break;
if(ev.size()>=2){
int x=ev.front();ev.pop();
int y=ev.front();ev.pop();
ans.push_back({x,y});
}
}
for(auto now:ans) printf("%d %d\n",now.first,now.second);
}
}
C. Number Game
题意：给你一个数n(<=1e9) 每次两个操作：
1、选择一个n的奇数的因子k，并将n代替为n/k
2、n减1
Ashishgup先手 FastestFinger后手，n为1的时候无法继续操作，谁无法操作谁就输了，输出赢的人名字
做法：简单博弈论，根据简单分析必胜态必败态：
（1必败） （2 必胜）（3必胜）（4必败）（5必胜）（6必败）
奇数必胜，偶数考虑唯一分解下，当2的次幂等于1 其他质数次幂大于1就是必胜态，比如2*3^2 我可以除3 变成2^3
当2的次幂大于1 其他质数次幂 之和大于0 就是必胜态。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=2e3+10;
int a[N],n;
int main()
{
int _=read();while(_--)
{
int n=read();
if(n==1){puts("FastestFinger");continue;}
if(n==2){puts("Ashishgup");continue;}
if(n%2){
puts("Ashishgup");continue;
}
int num1=0,num2=0;
for(int i=2;i*i<=n;++i){
if(n%i==0){
int res=0;
while(n%i==0) n=n/i,res++;
//printf("i:%d res:%d\n",i,res);
if(i==2) num1=res;
if(i%2) num2+=res;
}
}
if(n==2) num1++;
if(n!=1&&n%2) num2++;
//printf("num1:%d num2:%d\n",num1,num2);
if(num1==1){
if(num2>1) puts("Ashishgup");
else puts("FastestFinger");
}
else{
if(num2>0) puts("Ashishgup");
else puts("FastestFinger");
}
}
}
D. Odd-Even Subsequence
题意：给定n长度的序列a，以及k。要求从s中选取k长度的子序列使得以下这个值最小
做法：参考来自：博客
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=2e5+10;
int a[N],n,k;
int run(int mid)
{
int num0=k/2,num1=k-k/2;
int n0=0,n1=0;
for(int i=1;i<=n;++i){
if(a[i]<=mid) {
n1++;
if(i+1<=n) n0++,i++;
}
}
if(n1>=num1&&n0>=num0) return 1;
n1=n0=0;
n1=1;
for(int i=2;i<=n;++i){
if(a[i]<=mid) {
n0++;
if(i+1<=n) n1++,i++;
}
}
if(n1>=num1&&n0>=num0) return 1;
return 0;
}
int main()
{
n=read(),k=read();
rep(i,1,n) a[i]=read();
int mi=a[1],mx=a[1];
rep(i,2,n) mi=min(mi,a[i]),mx=max(mx,a[i]);
int l=mi,r=mx,ans;
while(l<=r){
int mid=l+r>>1;
if(run(mid)) ans=mid,r=mid-1;
else l=mid+1;
}
printf("%d\n",ans);
}
E. Binary Subsequence Rotation
题意和做法参考来自：博客
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N = 1000100;
char str1[N], str2[N];
int main()
{
int n, i, j, k, sum1, sum2, num1, num2;
sum1 = sum2 = 0;
scanf("%d %s %s", &n, str1, str2);
for(i=0;i<n;i++)sum1 += str1[i]-'0';
for(i=0;i<n;i++)sum2 += str2[i]-'0';
if(sum1 != sum2)printf("-1\n");
else{
num1 = num2 = 0;
for(i=0;i<n;i++)
if(str1[i] != str2[i]){
if(str1[i] == '0'){
if(num1 != 0)num1--;
num2++;
}else{
if(num2 != 0)num2--;
num1++;
}
}
printf("%d\n", num1+num2);
}
return 0;
}