Codeforces Round #564 (Div. 2) A到C D待补

#include <bits/stdc++.h>
using namespace std;
int main()
{
	int x,y,z;
	cin>>x>>y>>z;
	if(x>y+z){
		printf("+\n");
	}else if(y>x+z){
		printf("-\n");
	}else if(x==y&&z==0){
		printf("0\n");
	}else printf("?\n");
	
	return 0;
 } 

#include <bits/stdc++.h>
using namespace std;
int a[1001][1001];
int main()
{
	for(int i=1;i<=1000;i++)
		for(int j=1;j<=1000;j++)
			a[i][j] = i+j-2;
	int n,tn;
	cin>>n; tn = n - 1;
	if(tn%2==1)tn++;
	tn=(tn+2)/2;//tn行 
	printf("%d\n",tn);
	
	int cnt = 1;
	for(int i=1;i<=tn;i++)
		for(int j=1;j<=tn;j++){
			if(a[i][j]>=cnt-1){
				printf("%d %d\n",i,j);
				cnt++;
				if(cnt==n+1)break;
			}
			if(cnt==n+1)break;
		}
	
	return 0;
 } 

#include <bits/stdc++.h>
using namespace std;
struct No{
	int va,id;
}pipl[200005];
int hand[200005],xu[200005];
int n,tp,ans = 0,an;
bool judge2(){
	int tx = pipl[1].id,tf=1;
	an = tx - 1;
	int hand2[200005];
	for(int i=1;i<=n;i++)hand2[i] = hand[i];
	
	for(int i=tx+1;i<=n;i++){
		if(xu[i]!=xu[i-1]+1)return false;
	}
	tf = xu[n]+1; 
	int id=1;
	while(tf<=n){
		if(hand2[tf]==1){
			tf++;  if(id<tx)hand2[xu[id]]=1; id++;
		}else return false;
	}
	return true;
}
int main()
{
	cin.tie(0);
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>tp; hand[tp] = 1;
	}
	for(int i=1;i<=n;i++){
		cin>>xu[i];	pipl[xu[i]].id = i; pipl[xu[i]].va = xu[i];
	}
	if(judge2()){//判断是否可以直接插入 
		printf("%d\n",an);
		return 0; 
	}
	for(int i=1;i<=n;i++)
		if(hand[i]!=1){//不可直接插入则算出 最小最多无效用次数，即放0的个数 
			ans = max(ans,pipl[i].id - (i-1));
		}
	printf("%d\n",n+ans);
	return 0;
} 
/*
3
0 2 0
3 0 1
*/