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XCPC集训48

时间：2022-09-03 17:49:22浏览次数：83  
标签：48 int long read while XCPC include 集训 define

A - Square String?

/*
	qwq!
*/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <unordered_map>
using namespace std;
#define pb push_back
#define pu push
#define fi first
#define se second
#define LL long long
#pragma GCC optimize(2)
#define IOS ios::sync_with_stdio(false); std::cin.tie(0),cout.tie(0);
//#define int long long
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f, mod = 1e9 + 7;
const int N = 2e5 + 10;
//int h[N], e[N], ne[N], idx;
int w[N];


inline int read () {
	int k=0,f=1;
	char c=getchar ();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar ();}
	while (c>='0'&&c<='9') {k=k*10+c-'0';c=getchar ();}
	return k*f;
}

inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10-'0');
    return;
}


int main() {
	int T = read();
	while(T--) {
		string s; cin >> s;
		if(s.size() % 2) puts("NO");
		else {
			int x = s.size() / 2;
			if(s.substr(0, x) == s.substr(x)) puts("YES");
			else puts("NO");
		}
	}
	return 0;
}

B - Squares and Cubes

/*
	qwq!
*/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <unordered_map>
using namespace std;
#define pb push_back
#define pu push
#define fi first
#define se second
#define LL long long
#pragma GCC optimize(2)
#define IOS ios::sync_with_stdio(false); std::cin.tie(0),cout.tie(0);
#define int long long
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f, mod = 1e9 + 7;
const int N = 2e5 + 10;
//int h[N], e[N], ne[N], idx;
int w[N];
int cnt = 0;
unordered_map<int,int>mp;

inline int read () {
	int k=0,f=1;
	char c=getchar ();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar ();}
	while (c>='0'&&c<='9') {k=k*10+c-'0';c=getchar ();}
	return k*f;
}

inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10-'0');
    return;
}

inline bool check(int x) {
	bool success = false;
	int t = sqrt(x);
	if(t * t == x) return true;
	t = pow(x, 1.0 / 3.0);
	if(t*t*t == x) return true;
	return false;
}
signed main() {
	int T = read();
	while(T--) {
		int n = read();
		int ans = 0;
		mp.clear();
		for(int i = 1; i * i <= n; i++) {
			if(i * i <= n) mp[i * i]++;
			if(i * i * i <= n) mp[i * i * i] ++;
		}
		cout << mp.size() << endl;
	}	 
	return 0;
}

C - Wrong Addition

/*
	qwq!
*/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <unordered_map>
using namespace std;
#define pb push_back
#define pu push
#define fi first
#define se second
#define LL long long
#pragma GCC optimize(2)
#define IOS ios::sync_with_stdio(false); std::cin.tie(0),cout.tie(0);
//#define int long long
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f, mod = 1e9 + 7;
const int N = 2e5 + 10;
//int h[N], e[N], ne[N], idx;
int w[N];
int ans[N];


inline int read () {
	int k=0,f=1;
	char c=getchar ();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar ();}
	while (c>='0'&&c<='9') {k=k*10+c-'0';c=getchar ();}
	return k*f;
}

inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10-'0');
    return;
}


int main() {
	int T = read();
	while(T--) {
		string a, s; cin >> a >> s;
		int i = a.size() - 1, j = s.size() - 1;
		int k = 0;
		bool flag = true;
		while(j > -1 && i > -1) {
			int x = a[i] - '0';
			int y = s[j] - '0';
			if(y >= x) ans[k++] = y - x, j--, i--;
			else {
				if(!j) {
					flag = false;
					break;
				}
				j--;
			//	cout << y << "*" << endl;
				y = 10 * (s[j] - '0') + y;
			//	cout << y << endl;
				ans[k++] = y - x;
				if(y - x > 9 || y < x) {
					flag = false;
					break;
				}
				i--, j--;
			} 
		}
	//	cout << i << endl;
	//	cout << j << endl;
		if(!flag || i != -1) puts("-1");
		else {
			bool f = 0;
			while(j > -1) ans[k++] = s[j--] - '0';
			for(int i = k - 1; i >= 0; i--) {
				if(ans[i] > 0) f = 1;
				if(f) cout << ans[i];
			}
			cout << endl;
		}
	}
	return 0;
}

D - Not Shading

/*
	qwq!
*/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <unordered_map>
using namespace std;
#define pb push_back
#define pu push
#define fi first
#define se second
#define LL long long
#pragma GCC optimize(2)
#define IOS ios::sync_with_stdio(false); std::cin.tie(0),cout.tie(0);
//#define int long long
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f, mod = 1e9 + 7;
const int N = 110;
//int h[N], e[N], ne[N], idx;
char g[N][N];


inline int read () {
	int k=0,f=1;
	char c=getchar ();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar ();}
	while (c>='0'&&c<='9') {k=k*10+c-'0';c=getchar ();}
	return k*f;
}

inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10-'0');
    return;
}


int main() {
	int T = read();
	while(T--) {
		int n = read(), m = read(), r = read(), c = read();
		r--, c--;
		bool success = false;
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++) {
				cin >> g[i][j];
				if(g[i][j] == 'B') success = true;
			}
		if(!success) puts("-1");
		else {
			if(g[r][c] == 'B') {
				cout << 0 << endl;
				continue;
			}
			int r_cnt = 0, c_cnt = 0;
			for(int i = 0; i < m; i++) 
				if(g[r][i] == 'B') r_cnt++;
			for(int i = 0; i < n; i++) 
				if(g[i][c] == 'B') c_cnt++;
			int ans = 0;
			if(r_cnt > 0 || c_cnt > 0) ans = 1;
			else ans = 2;
			cout << ans << endl;
		}
		
	}
	return 0;
}

E - Not Sitting

思路：

因为A要尽量靠近B，B要尽量远离A，所以B一定会是在某个角落里，A就会趋近于整个图的中心，这样才能达到平衡。所以应当每次都涂里某个角落距离最近的点，且该点尽量靠近中心。

代码：

/*
	qwq!
*/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <unordered_map>
using namespace std;
#define pb push_back
#define pu push
#define fi first
#define se second
#define LL long long
#pragma GCC optimize(2)
#define IOS ios::sync_with_stdio(false); std::cin.tie(0),cout.tie(0);
//#define int long long
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f, mod = 1e9 + 7;
const int N = 2e5 + 10;
//int h[N], e[N], ne[N], idx;
int w[N];

inline int read () {
	int k=0,f=1;
	char c=getchar ();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar ();}
	while (c>='0'&&c<='9') {k=k*10+c-'0';c=getchar ();}
	return k*f;
}

inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10-'0');
    return;
}

int main() {
	int T = read();
	while(T--) {
		int n = read(), m = read();
		//int x = (n + 1) / 2, y = (m + 1) / 2;
		//cout << x << ' ' << y << endl;
		int k = 0;
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++) {
				w[k++] = max(i, n - i - 1) + max(j, m - j - 1);
			}
		sort(w, w + k);
	//	for(int i = 0; i < k; i++) cout << w[i].dist << endl;
	//	for(int i = 0; i < n * m - 1; i++) {
	//		int x = w[i].x + w[i].y - 2;
	//		cout << x << ' ';
	//	}
		for(int i = 0; i < k; i++) cout << w[i] << ' ';
		cout << endl;
	}
	return 0;
}

F - Triangle

思路：

斜着看每一行向上的个数，可得：(1 + 2^n) * (2^n) / 2, 要用到快速幂

代码：

/*
	qwq!
*/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <unordered_map>
using namespace std;
#define pb push_back
#define pu push
#define fi first
#define se second
#define LL long long
#pragma GCC optimize(2)
#define IOS ios::sync_with_stdio(false); std::cin.tie(0),cout.tie(0);
#define int long long
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f, mod = 1e9 + 7;
const int N = 2e5 + 10;
//int h[N], e[N], ne[N], idx;
int w[N];


inline int read () {
	int k=0,f=1;
	char c=getchar ();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar ();}
	while (c>='0'&&c<='9') {k=k*10+c-'0';c=getchar ();}
	return k*f;
}

inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10-'0');
    return;
}

inline LL qmi(LL a, LL b) {
	LL ans = 1;
	while(b) {
		if(b&1) ans = ans * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return ans % mod;
}

signed main() {
	LL n = read();
	if(n == 0) {
		cout << 1 << endl;
		return 0;
	}
	LL ans = (1 + qmi(2, n) % mod) * qmi(2, n - 1) % mod;
	cout << ans % mod << endl;
	return 0;
}

G - Banh-mi

思路：

对于某一段来说，肯定是优先选择1，最后1选完了再选0。 假设有a个1，b个0, 当作全是1，叠加总和为2^(a+b) - 1, 0拖了后腿，使得总和比实际多了2^b-1，所以实际为：
2(a+b)-1-(2b-1) = 2(a_b)-2b;
所以只需要统计区间0的个数就行了，可以用前缀和。

代码:

/*
	qwq!
*/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <unordered_map>
using namespace std;
#define pb push_back
#define pu push
#define fi first
#define se second
#define LL long long
#pragma GCC optimize(2)
#define IOS ios::sync_with_stdio(false); std::cin.tie(0),cout.tie(0);
#define int long long
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f, mod = 1e9 + 7;
const int N = 2e5 + 10;
//int h[N], e[N], ne[N], idx;
int w[N];


inline int read () {
	int k=0,f=1;
	char c=getchar ();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar ();}
	while (c>='0'&&c<='9') {k=k*10+c-'0';c=getchar ();}
	return k*f;
}

inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10-'0');
    return;
}

inline int qmi(int a, int b) {
	int ans = 1;
	while(b) {
		if(b & 1) ans = ans * a % mod;
		a = a * a % mod;
		b >>= 1; 
	}
	return ans % mod;
}
signed main() {
	int n = read(), m = read();
	string s; cin >> s;
	s = " " + s;
	for(int i = 1; i <= n; i++) {
		w[i] = w[i-1];
		if(s[i] == '0') w[i]++;
	} 
	while(m--) {
		int a = read(), b = read();
		int ans = (qmi(2, b - a + 1) - qmi(2, w[b] - w[a - 1])) % mod ;
		printf("%lld\n", (ans + mod) % mod);
	}
	return 0;
}

H - Anton and Fairy Tale

思路：

  • 当n<=m时，到第n天，仓库就空了。所以答案为n
  • 当n>m时：第m+1天开始数量才会开始减少：

m + 1天： n - (m + 1) + m = n - 1
m + 2天： n - 1 - (m + 2) = n - 1 - 2
...
所以可得第i天仓空： n <= i * (i + 1) / 2
故可以二分答案；

代码：

/*
	qwq!
*/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <unordered_map>
using namespace std;
#define pb push_back
#define pu push
#define fi first
#define se second
#define LL long long
#pragma GCC optimize(2)
#define IOS ios::sync_with_stdio(false); std::cin.tie(0),cout.tie(0);
#define int long long
//#define int __int64
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f, mod = 1e9 + 7, INT_MAX = 0x7fffffff;
const int N = 2e5 + 10;
//int h[N], e[N], ne[N], idx;
int w[N];


inline int read () {
	int k=0,f=1;
	char c=getchar ();
	while (c<'0'||c>'9') {
		if (c=='-') f=-1;
		c=getchar ();
	}
	while (c>='0'&&c<='9') {
		k=k*10+c-'0';
		c=getchar ();
	}
	return k*f;
}

inline void write(int x) {
	if(x<0) putchar('-'),x=-x;
	if(x>9) write(x/10);
	putchar(x%10-'0');
	return;
}


signed main() {
	int n = read(), m = read();
	if(n <= m) cout << n << endl;
	else {
		n -= m;
		int l = 0, r = 2e9;
		while(l < r) {
			int mid = l + r >> 1;
			//cout << mid << endl;
			if(mid * (mid + 1) / 2 >= n) r = mid;
			else l = mid + 1;
		}
		cout << m + r << endl;
	}
	return 0;
}

标签：48,int,long,read,while,XCPC,include,集训,define
From： https://www.cnblogs.com/MoonSkyy/p/16653150.html

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