2020计蒜客-蓝桥杯-模拟赛B组（三）题解

1.抛硬币
答案：0.50

2 杨辉三角

答案：92378

3.棋盘放置

答案：14
题解：在n*n的棋盘中，象最多可以摆 2 * n - 2 个

4 掷骰子

答案：126

5 旅行

答案：7
题解：每个点的方案使其他能到这个点的方案数之和。

6 剪刀石头布

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <map>
#include <queue>
#include <vector>
#include <cstring>
using namespace std;

const int N = 10005;
int a[5];


int main(){
	int n;  cin >> n;
	string s1,s2; cin >> s1 >> s2;
	for(int i = 0; i < n; i++){
		if(s1[i] - s2[i] == 0) a[1]++;
		else if(s1[i] == 'S'){
			if(s2[i] == 'R'){
				a[2]++;
			}else a[0]++;
		}else if(s1[i] == 'R'){
			if(s2[i] == 'S'){
				a[0]++;
			}else a[2]++;
		}else if(s1[i] == 'P'){
			if(s2[i] == 'S'){
				a[2]++;
			}else a[0]++;
		}
	}cout << a[0] << " " << a[2] << " " << a[1];
	return 0;
}

7 Chess

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <map>
#include <queue>
#include <vector>
#include <cstring>
using namespace std;

const int N = 10005;
int g[10][10]; 
int flag;

int jugheng(int x, int y){
	for(int i = x + 1; i <= 8; i++){ // xia
		if(g[i][y] != 0){
			return 1;
		}
	}
	for(int i = x - 1; i >= 0; i--){// shang
		if(g[i][y] != 0){
			return 1;
		}
	}
	for(int j = y + 1; j <= 8; j++){
		if(g[x][j] != 0){
			return 1;
		}
	}
	for(int j = y - 1; j >= 0; j--){
		if(g[x][j] != 0){
			return 1;
		}
	}return 0;
}

int jugcha(int x, int y){
	for(int i = x + 1, j = y + 1; i <= 8 && j <= 8; i++, j++){
		if(g[i][j] != 0) return 1;
	}
	for(int i = x - 1, j = y - 1; i >= 0 && j >= 0; i--, j--){
		if(g[i][j] != 0) return 1;
	}
	for(int i = x + 1, j = y - 1; i <= 8 && j >= 0; i++, j--){
		if(g[i][j] != 0) return 1;
	}
	for(int i = x - 1, j = y + 1; i >= 0 && j <= 8; i--, j++){
		if(g[i][j] != 0) return 1;
	}
}

void bfs(int x, int y){
	//cout << g[x][y] << endl;
	if(g[x][y] == 3){
		if(jugheng(x,y) == 1){
			flag = 1;
		}
	}else if(g[x][y] == 2){
		if(jugcha(x,y) == 1){
			flag = 1;
		}
	}else{
		if(jugheng(x,y) == 1 || jugcha(x,y) == 1){
			flag = 1;
		}
	}
}

int main(){
	int n; scanf("%d", &n);getchar();
	for(int i = 0; i < n; i++){
		char c; int x, y;
		scanf("%c%d%d", &c, &x, & y);getchar();
		if(c == 'Q'){ // huanghou 1
			g[x][y] = 1;
		}else if(c == 'B'){// xiang
			g[x][y] = 2;
		}else if(c == 'R'){// che
			g[x][y] = 3;
		}
		//cout << c << endl;
	}
	for(int i = 1; i <= 8; i++){
		for(int j = 1; j <= 8; j++){
			if(flag == 1){
				printf("Attack!");
				return 0;
			}
			if(g[i][j] != 0) bfs(i,j);
		}
	}if(flag == 1) printf("Attack!");
	else printf("Safe!");
	return 0;
}

/*
2
R 1 1
R 2 1
*/

/*
2
R 1 1
R 2 2
*/

/*
2
B 1 1
B 2 1
*/

/*
2
B 1 1
B 2 2
*/

8 分披萨

#include <iostream>
#include <algorithm>
using namespace std;

const int N = 100005;

int n,m;
int a[N];

bool cmp(int a, int b){
	return a > b;
}


int cal(int x){
	int sum = 0;
	for(int i = 0; i < m; i++){
		if(a[i] / x == 0) break;
		sum += (a[i] / x);
	}if(sum >= n) return 1;
	return 0;
}


int main(){
	scanf("%d%d", &n, &m);
	for(int i = 0; i < m; i++){
		scanf("%d", &a[i]);
	}sort(a, a + m, cmp);
	int l = 1, r = N;
	while(l < r){
		int mid = (l + r) / 2;
		if(cal(mid) == 0){
			r = mid;
		}else{
			l = mid + 1;
		}
	}cout << --l;
	return 0;
}
/*
4 1
4
*/

/*
1 1
4
*/ 

/*
4 2
4 5
*/

 

8 养猫


**题解：**构造哈夫曼树，这里可以用优先队列来解决

#include <iostream>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;

priority_queue<int, vector<int>, greater<int> > q;

int main(){
	int n;scanf("%d", &n);
	for(int i = 0; i < n; i++){
		int t;scanf("%d", &t); q.push(t);
	}int sum = 0;
	while(q.size() > 1){
		int a = q.top(); q.pop();
		int b = q.top(); q.pop();
		sum += (a + b);
		q.push((a + b));
	}printf("%d", sum);
	return 0;
} 

10 突破障碍


题解：利用广搜配合双端队列去找，碰到#放到后端, .和T放到前端，然后最先找到的就是用步长最少的，其中步长指的是#的个数

#include <iostream>
#include <cstring>
#include <deque>
#include <algorithm>
#include <cstdio>
using namespace std;

const int N = 305;

int n,m;
char g[N][N];
int bi,bj,ei,ej;
int vis[N][N];
int ans;
int dx[5] = {0, 0, 1, -1};
int dy[5] = {1 , -1, 0, 0};

struct node{
	int x,y;
	int cnt;
};

deque<node> q;

void bfs(){
	q.push_back(node{bi,bj,0});
	node t;
	while(q.size()){
		t = q.front(); q.pop_front();
		if(t.x == ei && t.y == ej){
		//	cout << t.cnt << endl;
			ans = t.cnt;
		//	cout << ans << endl;
			return;
		}
		vis[t.x][t.y] = 1;
		
		for(int i = 0; i < 4; i++){
			int x = t.x + dx[i];
			int y = t.y + dy[i];
			if(x >= 0 && y >= 0 && x < n && y < m && vis[x][y] == 0 ){
				if(g[x][y] == '.' || g[x][y] == 'T'){
					q.push_front(node{x, y, t.cnt});
				}else{
					q.push_back(node{x,y, t.cnt + 1});
				}
			}
		}
	}
}


int main(){
	scanf("%d%d", &n, &m); getchar();
	for(int i = 0; i < n; i++){
		for(int j = 0; j < m; j++){
			scanf("%c", &g[i][j]);
			if(g[i][j] == 'S') bi = i, bj = j;
			if(g[i][j] == 'T') ei = i, ej = j;
		}getchar();
	}
	bfs();
	printf("%d", ans);	
	return 0;
}
``