Link~
题面差评，整场都在读题
A
根据奇偶性判断一下即可。
#include<bits/stdc++.h>
#define ll long long
#define N
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define lowbit(i) ((i)&(-i))
#define VI vector<int>
using namespace std;
int t,n;
string s;
int main(){
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
cin>>t;
while(t--){
cin>>n;
cin>>s;
int odd,even;odd = even = -1;
if(n&1){
rep(i,0,n-1){
if((i&1)==0){
int v = s[i]-'0';
if(v&1) odd = 1;
else even = 1;
}
}
if(odd != -1) puts("1");
else puts("2");
}else{
rep(i,0,n-1){
if((i&1)==1){
int v = s[i]-'0';
//cout << "v: " <<v << endl;
if(v&1) odd = 1;
else even = 1;
}
}
if(even != -1) puts("2");
else puts("1");
}
}
return 0;
}
B
神必题意，读了快半小时才明白题意。
看着题目给的图的找个规律就行了。
#include<bits/stdc++.h>
#define int long long
#define N 1000015
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define lowbit(i) ((i)&(-i))
#define VI vector<int>
using namespace std;
int t,n,x,val[N];
int qpow(int a,int b){
int res = 1;
while(b){
if(b&1) res = res*a;
a = a*a;
b>>=1;
}
return res;
}
signed main(){
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
cin>>t;n = 31;
rep(i,1,n) val[i] = val[i-1]*2+qpow(4,i-1);
rep(i,1,n) val[i] += val[i-1];//,cout << val[i] << endl;
while(t--){
cin>>x;
int l = 1,r = n,ans = 0;
while(l+3 < r){
int mid = (l+r)>>1;
if(val[mid] <= x) l = mid;
else r = mid;
}
rep(i,l,r){
if(val[i] <= x) ans = max(ans,i);
}
cout << ans << endl;
}
return 0;
}
C
大力分类讨论题，WA了好多发
#include<bits/stdc++.h>
#define int long long
#define N 1005
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define lowbit(i) ((i)&(-i))
#define VI vector<int>
using namespace std;
int t,n,a[N];
signed main(){
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
cin>>t;
while(t--){
int ans = inf,cur,ff = 0;
cin>>n>>cur;
rep(i,1,n) cin>>a[i];
int sum = 0;rep(i,1,n) sum += a[i];
rep(i,1,n) if(a[i] == cur) ff++;
if(n*cur == sum) ans = min(ans,1ll);
if(ff == n) ans = min(ans,0ll);
if(ff > 1){
ans = min(ans,1ll);
}else{
ans = min(ans,2ll);
}
int left = n-ff;
if(left < n) ans = min(ans,1ll);
if(sum%n == 0){
if(ff) ans = min(ans,1ll);
}
cout << ans << endl;
}
return 0;
}
D
贪心构造
#include<bits/stdc++.h>
#define ll long long
#define N 100015
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define lowbit(i) ((i)&(-i))
#define VI vector<int>
using namespace std;
int n,a[N],b[N];
int main(){
//freopen(".in","r",stdin);
//freopen(".out","w",stdout);
cin>>n;
rep(i,1,n) cin>>a[i];
sort(a+1,a+n+1);int l = 1,r = n/2+1;
rep(i,1,n){
if(i&1) b[i] = a[r++];
else b[i] = a[l++];
}
int cnt = 0;
rep(i,2,n-1){
if(b[i] < b[i-1]&&b[i] < b[i+1]) cnt++;
}
cout << cnt << endl;
rep(i,1,n) cout << b[i] << ' ';
return 0;
}
E
特判\(n = p*q\)的情况，其中\(p,q\)是质数，答案为\(1\)，直接输出即可。
否则答案为零，
这样围成一个圈就行了。
垃圾场，让人挺不爽的，因为奇怪的原因被区分，可能是手速和英语不太行。