"科林明伦杯"哈尔滨理工大学第八届程序设计竞赛 题解

题目链接  Problems

Problem A

快速幂累加即可。

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

LL ans = 0;
LL n, d;
int T;

const LL mod = 1e9 + 7;

inline LL Pow(LL a, LL b, LL Mod){
        LL ret(1);
        for (; b; b >>= 1, (a *= a) %= Mod) if (b & 1) (ret *= a) %= Mod;
        return ret;
}


int main(){

	scanf("%d", &T);
	while (T--){
		cin >> n >> d;
		ans = 0;
		rep(i, 1, n){
			ans += Pow(i, d, mod);
			ans %= mod;
		}

		printf("%lld
", ans);
	}

	return 0;
}

Problem B

对于每个帮派，并查集维护就可以了。

求第$k$大的时候树状数组上二分就好了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

const int N = 1e5 + 10;

int T;
int sz[N], c[N], father[N];
int n, m;
int num;

void update(int x, int val){
	for (; x <= n; x += x & -x) c[x] += val;
}

int query(int x){
	int ret = 0;
	for (; x; x -= x & -x) ret += c[x];
	return ret;
}

int getfather(int x){
	return father[x] == x ? x : father[x] = getfather(father[x]);
}


int main(){

	scanf("%d", &T);
	while (T--){
		scanf("%d%d", &n, &m);
		memset(c, 0, sizeof c);
		memset(father, 0, sizeof father);
		rep(i, 1, n) father[i] = i;

		rep(i, 1, n) update(1, 1);
		num = n;
		rep(i, 1, n) sz[i] = 1;
		while (m--){
			int op;
			scanf("%d", &op);
			if (op == 1){
				int x, y;
				scanf("%d%d", &x, &y);
				int fx = getfather(x), fy = getfather(y);
				if (fx == fy) continue;

				father[fy] = fx;

				int f1 = sz[fx], f2 = sz[fy], f3 = sz[fx] + sz[fy];
				sz[fx] += sz[fy];
				sz[fy] = 0;
				--num;

				update(f1, -1);
				update(f2, -1);
				update(f3, 1);
			}

			else{
				int x;
				scanf("%d", &x);
				if (num < x){
					puts("-1");
					continue;
				}
				int l = 1, r = n;
				
				while (l + 1 < r){
					int mid = (l + r) >> 1;
					if (num - query(mid - 1) >= x) l = mid;
					else r = mid - 1;
				}

				if (num - query(r - 1) >= x) printf("%d
", r);
				else printf("%d
", l);
			}
		}
	}

	return 0;
}

Problem C

递推。$f_{n} = 2f_{n-1} + f_{n-3}$

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

const int A = 5;

const LL mod = 1e9 + 7;

struct matrix{ LL a[A][A];}  init, unit, aa;
int n;
LL m;
int T;

matrix Mul(matrix a, matrix b){
	matrix c;
	rep(i, 1, n) rep(j, 1, n){
		c.a[i][j] = 0;
		rep(k, 1, n) (c.a[i][j] += (a.a[i][k] * b.a[k][j] % mod)) %= mod;
	}
	return c;
}

matrix Pow(matrix a, LL k){
	matrix ret(unit); for (; k; k >>= 1ll, a = Mul(a, a)) if (k & 1) ret = Mul(ret, a); return ret;
}	



int main(){

	n = 3;

	matrix dd;
	memset(dd.a, 0, sizeof dd.a);

	memset(unit.a, 0, sizeof unit.a);
	rep(i, 1, n) unit.a[i][i] = 1ll;

	dd.a[1][1] = 2; dd.a[1][2] = 0;  dd.a[1][3] = 1;
	dd.a[2][1] = 1;
	dd.a[3][2] = 1;

	scanf("%d", &T);
	while (T--){
		scanf("%lld", &m);

		LL fuck = m - 3;
		if (m <= 3){
			if (m == 1ll) puts("1");
			if (m == 2ll) puts("2");
			if (m == 3ll) puts("5");
			continue;
		}

		matrix cc = Pow(dd, fuck);
		
		LL ans = cc.a[1][1] * 5ll + cc.a[1][2] * 2ll + cc.a[1][3] * 1ll;
		ans %= mod;

		printf("%lld
", ans);
	}

	return 0;
}

Problem D

最坏的情况即为斐波那契数列中的某几项。

那么当询问元素个数超过一定的时候（大概$87$）直接输出Yes就好了。

否则就暴力特判。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

LL a[100010];
LL c[100010];
int n, q;
int cnt = 0;


int main(){

	scanf("%d", &n);
	rep(i, 1, n) scanf("%lld", a + i);

	scanf("%d", &q);
	while (q--){
		int l, r;
		scanf("%d%d", &l, &r);
		if (r - l + 1 >= 100){
			puts("Yes");
			continue;
		}
		cnt = 0;
		rep(i, l, r) c[++cnt] = a[i];

		sort(c + 1, c + cnt + 1);


		int fg = 0;
		rep(i, 1, cnt - 2) if (c[i] + c[i + 1] > c[i + 2]){
			fg = 1;
			break;
		}

		puts(fg ? "Yes" : "No");
	}

	return 0;
}

Problem E

分解质因数之后令$a_{i}$为每个质因数的指数。

答案为

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

LL n;
int T;
LL bs;
LL cc;
LL ans;


int main(){

	while (~scanf("%d", &T)){
		while (T--){
			scanf("%lld", &n);
			LL bs = sqrt(n);

			ans = 1;

			for (LL i = 2; i <= sqrt(n); ++i){
				LL cc = 0;
				while (n % i == 0) ++cc, n /= i;
				ans *= (2 * cc + 1ll);
			}

			if (n > 1) ans *= 3;
			++ans;
			ans /= 2;
			printf("%lld
", ans);
		}
	}

	return 0;
}

Problem F

答案为$2^{n-3} * n^{2} * (n+3)$

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

const LL mod = 1e9 + 7;

int T;
LL n, ans;

inline LL Pow(LL a, LL b, LL Mod){
        LL ret(1);
        for (; b; b >>= 1, (a *= a) %= Mod) if (b & 1) (ret *= a) %= Mod;
        return ret;
}

int main(){

	scanf("%d", &T);
	while (T--){
		scanf("%lld", &n);
		if (n == 1ll){
			puts("1");
			continue;
		}

		if (n == 2ll){
			puts("10");
			continue;
		}

		if (n == 3ll){
			puts("54");
			continue;
		}

		LL ans = Pow(2, n - 3, mod);
		ans *= n;
		ans %= mod;
		ans *= n;
		ans %= mod;
		ans *= (n + 3ll);
		ans %= mod;

		printf("%lld
", ans);
	}


	return 0;
}

Problem G

从小到大枚举答案，每次做一遍极大极小搜索，若符合题意就直接输出。

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

const int N = 2e3 + 10;

LL a[N];
LL f[2][N];
LL s[N];
LL xxx;
int n;
int m;
int ans;

LL dp(int x, int pos){
	if (~f[x][pos]) return f[x][pos];

	if (pos == m){
		if (x) return f[x][pos] = a[pos];
		else return f[x][pos] = 0;
	}

	LL ret = 0;
	if (x){	
		ret = a[pos] + dp(x ^ 1, pos + 1);
		ret = max(ret, dp(x, pos + 1));
	}

	else{
		ret = dp(x ^ 1, pos + 1);
		ret = min(ret, a[pos] + dp(x, pos + 1));
	}

	return f[x][pos] = ret;
}

int main(){

	while (~scanf("%d", &n) && n != -1){
		rep(i, 1, n) scanf("%lld", a + i);
		scanf("%lld", &xxx);
		s[0] = 0;
		rep(i, 1, n) s[i] = s[i - 1] + a[i];
		ans = -1;
		for (m = 1; m <= n; ++m){
			memset(f, -1, sizeof f);
			LL gg = dp(1, 1);
			if (gg >= xxx){
				ans = m;
				break;
			}
		}

		printf("%d
", ans);			
	}

	return 0;
}

Problem H

贪心。求出每个块的大小，然后枚举每个块。记块的个数为$cnt$

两边的块如果有不小于$2$的，那么答案用$cnt + 1$更新。

中间的块大小如果有不小于$3$的，那么答案用$cnt + 2$更新。

UPD：哦草我好像没考虑0011然后翻转中间的0和1的情况，这也是一个case

代码就不改乐

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

const int N = 1e5 + 10;

int a[N];
int c[N];
int n, cnt, xx, now;
int ans;


int main(){

	while (~scanf("%d", &n)){
		rep(i, 1, n) scanf("%1d", a + i);

		xx = -1;
		cnt = -1; now = 0;
		rep(i, 1, n){
			if (a[i] != xx){
				c[++cnt] = now;
				now = 1;
			}

			else ++now;

			xx = a[i];
		}


		c[++cnt] = now;
		ans = cnt;
		if (c[1] == 2 || c[cnt] == 2) ans = max(ans, cnt + 1);
		rep(i, 1, cnt) if (c[i] >= 3) ans = max(ans, cnt + 2);
		printf("%d
", ans);
	}

	return 0;
}

Problem I

模拟题。

#include <cstdio>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second

typedef long long LL;

char s[11010];
int T;
set <string> mp;
int l;
int n;
set <string> :: iterator it;

int judge(char ch){
	if ((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z') || (ch >= '0' && ch <= '9') || (ch == '-') || (ch == '_')) return 1;
	return 0;
}


int main(){

	scanf("%d", &T);
	while (T--){
		mp.clear();
		scanf("%d", &n);
		getchar();

		rep(op, 1, n){
			gets(s);
			l = strlen(s);

			string s1 = "";
			int i;
			for (i = 0; i < l; ++i){
				if (s[i] == '@'){
					if (i && judge(s[i - 1])) continue;
					s1 = "";
					for (; i + 1 < l && judge(s[i + 1]); ){
						s1 += s[i + 1];
						++i;
					}

					if (s1 != "") mp.insert(s1);
				}
			}


		}

		printf("%d
", (int)mp.size());
		for (it = mp.begin(); it != mp.end(); ++it) cout << *it << endl;
	}


	return 0;
}

Problem J

签到。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

int a[100010], b[100010];
int n ;
int T;


int main(){

	scanf("%d", &T);
	while (T--){
		scanf("%d", &n);
		rep(i, 1, n) scanf("%d", a + i);
		rep(i, 1, n) scanf("%d", b + i);

		int ff = 1, fg = 1;

		rep(i, 1, n) if (a[i] > b[i]) ff = 0;
		rep(i, 1, n) if (a[i] > b[n - i + 1]) fg = 0;

		if (ff && fg) puts("both");
		else if (ff && !fg) puts("front");
		else if (!ff && fg) puts("back");
		else puts("none");
	}


	return 0;
}