Codeforces Round #829 (Div. 2) A~D

比赛链接：Dashboard - Codeforces Round #829 (Div. 2) - Codeforces

A. Technical Support

B. Kevin and Permutation

C1. Make Nonzero Sum (easy version)

C2. Make Nonzero Sum (hard version)

D. Factorial Divisibility

A. Technical Support

题意：给定一个由QA组成的字符串，Q表示问题，A表示回答，要求每一个Q后面都能对应至少一个A。

思路：从后往前遍历，遇到‘Q’的时候判断‘A’的个数是否大于‘Q’，若不是则输出NO。

#define _CRT_SECURE_NO_WARNINGS 1
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<math.h>
#define FOR(a,b) for(int i=a;i<=b;i++)
#define ROF(a,b) for(int i=a;i>=b;i--)
#define FORj(a,b) for(int j=a;j<=b;j++)
#define ROFj(a,b) for(int j=a;j>=b;j--)
#define FORk(a,b) for(int k=a;k<=b;k++)
#define ROFk(a,b) for(int k=a;k>=b;k--)
#define mem(i,a) memset(i,a,sizeof(i))
#define ll long long
#define inf 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define int  long long
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define pi acos(-1.0)
#define endl '\n'
using namespace std;
const int maxn = 5e6 + 5;
void solve() {int n,A=0,Q=0;string s;cin >> n >> s;ROF(n - 1, 0) {if (s[i] == 'A')A++;else Q++;if (s[i] == 'Q' && A < Q) {no;return;}}yes;
}
signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int _;cin >> _;while (_--)solve();return 0;
}

B. Kevin and Permutation

题意：给定数组长度n，请构造一个排列使得两个相邻元素差的最小值最大。

思路：我们可以俩俩一组构造，像n/2+1, 1, n/2 + 2, 2......这样子，若数组长度为奇数，我们只用将n放到数组最后即可。

#define _CRT_SECURE_NO_WARNINGS 1
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<math.h>
#define FOR(a,b) for(int i=a;i<=b;i++)
#define ROF(a,b) for(int i=a;i>=b;i--)
#define FORj(a,b) for(int j=a;j<=b;j++)
#define ROFj(a,b) for(int j=a;j>=b;j--)
#define FORk(a,b) for(int k=a;k<=b;k++)
#define ROFk(a,b) for(int k=a;k>=b;k--)
#define mem(i,a) memset(i,a,sizeof(i))
#define ll long long
#define inf 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define int long long
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define pi acos(-1.0)
#define endl '\n'
using namespace std;
const int maxn = 2e5 + 5;
void solve() {int n;cin >> n;FOR(1,n/2)  cout << i + n / 2 <<" " << i << " ";if(n%2)cout<<n;cout<<endl;
}
signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int _;cin >> _;while (_--)solve();return 0;
}

C1. Make Nonzero Sum (easy version)

题意：给定一个[-1, 1]组成的数组，将该数组拆分成若干区间，每一个区间的价值为a[l]-a[l+1]+a[l+2]-a[l+3]......a[r]。需要我们构造一个划分区间的方法使得总价值为0。

思路：若1和-1的总数目为奇数则构造不了使得总价值为0的方法，因为划分区间在本质上只是能吧一些-1变成1，把一些1变成-1，或者不变。所以每次总价值只会-2或者+2，若n为奇数那么初始的总价值也为奇数，改变区间对应改变的总价值也只会是偶数，那么总价值永远到不了0。

当n为偶数时，我们可以两两分组，每组选择两次长度为1的区间，或一次长度为2的区间，若选择两次长度为1的区间会使得总价值不变，选择一次长度为2的区间会使得总价值-2或者总价值+2，我们最多选择n/2组，所以当n为偶数的时候必有解。

ps：C题我写的代码比较乱，建议看别人的代码或者自己写。

#define _CRT_SECURE_NO_WARNINGS 1
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<math.h>
#define FOR(a,b) for(int i=a;i<=b;i++)
#define ROF(a,b) for(int i=a;i>=b;i--)
#define FORj(a,b) for(int j=a;j<=b;j++)
#define ROFj(a,b) for(int j=a;j>=b;j--)
#define FORk(a,b) for(int k=a;k<=b;k++)
#define ROFk(a,b) for(int k=a;k>=b;k--)
#define mem(i,a) memset(i,a,sizeof(i))
#define ll long long
#define inf 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define int long long
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define pi acos(-1.0)
#define endl '\n'
using namespace std;
const int maxn = 2e5 + 5;
int a[maxn];
pair<int, int>b[maxn];
void solve() {int n, z = 0, f = 0, k = 0;cin >> n;FOR(1, n) {cin >> a[i];if (a[i] == 1)z++;else f++;}if (z == f) {cout << n << endl;FOR(1, n)cout << i << " " << i << endl;return;}if (n % 2) {cout << -1 << endl;return;}if (z > f) {FOR(1, n - 1) {if (a[i + 1] == 1) {b[++k].first = i, b[k].second = i + 1;z--;f++;i++;} else   b[++k].first = i, b[k].second = i;if (z == f)break;}} else {FOR(1, n - 1) {if (a[i + 1] == -1) {b[++k].first = i, b[k].second = i + 1;f--;z++;i++;} else  b[++k].first = i, b[k].second = i;if (z == f)break;}}cout << k + n - b[k].second << endl;FOR(1, k)cout << b[i].first << " " << b[i].second << endl;FOR(b[k].second + 1, n)cout << i << " " << i << endl;
}
signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int _;cin >> _;while (_--)solve();return 0;
}

C2. Make Nonzero Sum (hard version)

题意：给定一个[-1, 0,1]组成的数组，将该数组拆分成若干区间，每一个区间的价值为a[l]-a[l+1]+a[l+2]-a[l+3]......a[r]。需要我们构造一个划分区间的方法使得总价值为0。(和C1相比就是数组中除了1和-1多了0)

思路：和C1大同小异，也是两个不为0的数组两个为一组，我们需要先获取两个不为0的数字的下标，就可以构造出每组中第二个数字取反的区间方法了，同样每次操作能使总价值+2或者-2。

#define _CRT_SECURE_NO_WARNINGS 1
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<math.h>
#define FOR(a,b) for(int i=a;i<=b;i++)
#define ROF(a,b) for(int i=a;i>=b;i--)
#define FORj(a,b) for(int j=a;j<=b;j++)
#define ROFj(a,b) for(int j=a;j>=b;j--)
#define FORk(a,b) for(int k=a;k<=b;k++)
#define ROFk(a,b) for(int k=a;k>=b;k--)
#define mem(i,a) memset(i,a,sizeof(i))
#define ll long long
#define inf 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define int long long
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define pi acos(-1.0)
#define endl '\n'
using namespace std;
const int maxn = 2e5 + 5;
int a[maxn];
pair<int, int>b[maxn];
void solve() {int n, z = 0, f = 0, k = 0;cin >> n;FOR(1, n) {cin >> a[i];if (a[i] == 1)z++;else if(a[i]==-1)f++;}if (z == f) {cout << n << endl;FOR(1, n)cout << i << " " << i << endl;return;}if ((z+f)% 2) {cout << -1 << endl;return;}int p1= 1,p2=2;if (z > f) {while(p2<=n) {while (!a[p2] && p2 <= n)p2++;if (a[p2] == 1) {if ((p2 - p1) % 2)b[++k].first = p1, b[k].second = p2;else {b[++k].first = p1, b[k].second = p1;b[++k].first = p1 + 1, b[k].second = p2;}z--, f++;} else {if ((p2 - p1) % 2) {b[++k].first = p1, b[k].second = p1;b[++k].first = p1 + 1, b[k].second = p2;} else {b[++k].first = p1, b[k].second = p2;}}if (z == f)break;p1 = p2 + 1;p2=p1+1;}} else {while (p2 <= n) {while (!a[p2] && p2 <= n)p2++;
//           cout<<p1<<" "<<p2<<" "<<a[p1]<<" "<<a[p2]<<endl;if (a[p2] == -1) {if ((p2 - p1) % 2)b[++k].first = p1, b[k].second = p2;else {b[++k].first = p1, b[k].second = p1;b[++k].first = p1 + 1, b[k].second = p2;}z++, f--;} else {if ((p2 - p1) % 2) {b[++k].first = p1, b[k].second = p1;b[++k].first = p1 + 1, b[k].second = p2;} else {b[++k].first = p1, b[k].second = p2;}}if (z == f)break;p1 = p2 + 1;p2=p1+1;}}cout << k + n - b[k].second << endl;FOR(1, k)cout << b[i].first << " " << b[i].second << endl;FOR(b[k].second + 1, n)cout << i << " " << i << endl;}
signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int _;cin >> _;while (_--)solve();return 0;
}

D. Factorial Divisibility

题意：给定一个长度为n的数组和k，判断!k是否整除(!a1 + !a2 + !a3 + !a4... + !an)。

思路：我们可以发现一个规律：1！+1！=2！，2！+2！+2！=3！，3！+3！+3！+3！=4！......

所以可得出n个x！可以转化为n/(x+1)个（x+1）!，所以我们从低位到高位累加，利用mp[i] % (i + 1)是否等于0，来判断每位能不能将数字全部转化，若不能被全部转化就说明了不能被整除。

#define _CRT_SECURE_NO_WARNINGS 1
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<math.h>
#define FOR(a,b) for(int i=a;i<=b;i++)
#define ROF(a,b) for(int i=a;i>=b;i--)
#define FORj(a,b) for(int j=a;j<=b;j++)
#define ROFj(a,b) for(int j=a;j>=b;j--)
#define FORk(a,b) for(int k=a;k<=b;k++)
#define ROFk(a,b) for(int k=a;k>=b;k--)
#define mem(i,a) memset(i,a,sizeof(i))
#define ll long long
#define inf 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define int unsigned long long
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define pi acos(-1.0)
#define endl '\n'
using namespace std;
const int maxn = 5e6 + 5;
map<int, int>a;
void solve() {int n, k;cin >> n >> k;FOR(1, n) {int x;cin >> x;a[x]++;}FOR(1, k-1) {if (a[i] % (i + 1)) {no;return;}a[i + 1] += a[i] / (i + 1);}yes;
}
signed main() {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);//    int _;//    cin >> _;//   while (_--)solve();return 0;
}