Codeforces Round #807 (Div. 2) A--D

A

思路：排序之后，就将数据分割成前n和后n就行，

12345678->

5678

1234

这样一一对应相差值最大

/**/
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = a; i <= b; i ++)
#define endl '\n'
using namespace std;
const int N = 500 + 5;
 
void s(){
  int n ,x; cin >> n >> x;
  
  int a[N] = {}, a1[N] = {}, b1[N] = {};
  for(int i = 1; i <= 2 * n ;i ++) cin >> a[i];
  
  sort(a+1, a+1+n*2);
  
  
  for(int i = 1; i <= n ;i ++) a1[i] = a[i];
  rep(i, n + 1, 2*n) b1[i - n] = a[i]; //升序 b1是高的
  
  for(int i = 1; i <= n; i++){
    if(b1[i] - a1[i] < x) {
      cout << "NO\n";return ;
    }
  }
  
  cout << "YES\n";
  
  
}
 
int main(){
 
  cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
 
  #ifdef LOCAL
  cout << "记得更改input.txt!!!\n";
  FILE *p = fopen("input.txt", "a+");
  fclose(p);
  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);
  #endif
  int t; cin >> t; while(t-- )s();
 
 
  return 0;
}
 
//提交代码删掉#define LOCAL

B

思路：

将一堆沙子往后移动需要有铺垫。

前缀和+（去掉前导0，如果最后a[n]是0也要去掉）0的个数 = 答案

/**/
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = a; i <= b; i ++)
#define endl '\n'
using namespace std;
const int N = 2e5 + 5;
long long sum[N] = {};
void s(){
  int n ; cin >> n;
  
  int a[N] = {};
  memset(sum, 0, sizeof sum);
  rep(i,1,n) cin >> a[i];
  
  bool tag = 0;
  int cnt  = 0;
  int id = 1;
  while(a[id] == 0) id++;
  rep(i,id,n){
    sum[i] = sum[i - 1] + a[i];
    if(a[i] == 0) cnt++;
  }
  if(a[n] == 0 && cnt != 0) cnt --;
  
  
  cout << sum[n - 1] + cnt << endl;
  
  
}
 
int main(){
 
  cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);
 
  #ifdef LOCAL
  cout << "记得更改input.txt!!!\n";
  FILE *p = fopen("input.txt", "a+");
  fclose(p);
  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);
  #endif
  int t; cin >> t;while(t--) s();
  
  
  return 0;
}
 
//提交代码删掉#define LOCAL

C

思路：先记录每一个区间。

每一次询问都倒着推导k所在的位置。最终递推出k在1--2e5之间的值（也就是在原数据s中的值）

/**/
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = a; i <= b; i ++)
#define endl '\n'
using namespace std;
const int N = 2e5 + 5;
char s[N];
long long l[50], r[50];//c=40
long long sum[50];

void ss(){
  int n , c , q;
  cin >> n >> c >> q >> s + 1;
  sum[0] = n;
  for(int i = 1;i <= c;i  ++){
    cin >> l[i] >> r[i];//记录复制的区间
    sum[i] = sum[i - 1] + r[i] - l[i] + 1;//sum[i]是当前多长了
  }
  
  while(q--){
    long long k;
    cin >> k;
    
    for(int i = c; i >= 1; i --){//每一次询问都倒着查询。
      if(k > sum[i - 1] && k <= sum[i]){//如果k落在前缀和区间内
        k = l[i] + k - sum[i - 1] - 1;//定位到k的位置，循环不断的缩小k的值，最终k会落在1--2e5的范围，答案就是s[k]
      }
    }
    cout << s[k] << endl;
  }
}

int main(){

  cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);

  #ifdef LOCAL
  cout << "记得更改input.txt!!!\n";
  FILE *p = fopen("input.txt", "a+");
  fclose(p);
  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);
  #endif
  
  
  int t; cin >> t; while(t--) ss();
  
  
  return 0;
}

//提交代码删掉#define LOCAL

D

/**/
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = a; i <= b; i ++)
#define endl '\n'
using namespace std;
const int N = 2e5 + 5;

char s1[N], s2[N];

void s(){
  int n ; cin >> n >> s1 + 1 >> s2 + 1;
  if(s1[1] != s2[1] || s1[n] != s2[n]){//判断两头是否相等，不相等是永远无法处理的
    cout << -1 << endl;
    return ;
  }
  
  
  vector<pair<int, int>> v1, v2;
  for(int i = 1; i <= n ;i ++){
    if(s1[i] == '0') continue;
    int j = i ;
    while(j + 1 <= n && s1[j + 1] == '1') j ++;
    v1.push_back({i, j});
    i = j;
  }
  
  for(int i = 1; i <= n ;i ++){
    if(s2[i] == '0') continue;
    int j = i;
    while(j + 1 <= n && s2[j + 1] == '1') j++;
    v2.push_back({i, j});
    i = j;
  }
  
  if(v1.size() != v2.size()){
    cout << -1 << endl; return;
  }
  
  long long ans = 0;
  for(int i = 0 ; i < v1.size(); i++){
    pair<int, int> tmp1, tmp2;
    tmp1 = v1[i];
    tmp2 = v2[i];
    int l1 = tmp1.first, r1 = tmp1.second, l2 = tmp2.first, r2 = tmp2.second; 

    ans += abs(l2 - l1) + abs(r2 - r1);
  }
  
  cout << ans << endl;
}

int main(){

  cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);

  #ifdef LOCAL
  cout << "记得更改input.txt!!!\n";
  FILE *p = fopen("input.txt", "a+");
  fclose(p);
  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);
  #endif
  int t; cin >> t;
  while(t--) s();
  
  
  return 0;
}

//提交代码删掉#define LOCAL