2019/08/18 练习赛

T1

思路

换算成相同的单位直接相加减

代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std ;
int a , b ;
int con = 19 ;
int main () {
	freopen("pencil.in","r",stdin) ;
	freopen("pencil.out","w",stdout) ;
	cin >> a >> b ;
	int ans = a*10+b ;
	cout << floor(ans / con) <<endl ;
	return 0 ;
} 

T2

思路

打表找规律,容易发现答案就是(2^{n-1})
至于(2^{n-1})的二进制emmmm反正我不会(手动滑稽)

code

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std ;
int T , n ; 
int main () {
	freopen("water.in","r",stdin) ;
	freopen("water.out","w",stdout) ;
	scanf("%d",&T) ;
	while(T --) {
		scanf("%d",&n) ;
		printf("1") ;
		for(int i = 1 ; i < n ; i ++) {
			printf("0") ;
		}
		puts("") ;
	}
	return 0 ;
}

T3

思路

只要别想多就是到比较不错的好模拟

code

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std ;
int n , A , ans ;
int main () {
	freopen("Weita.in","r",stdin) ;
	freopen("Weita.out","w",stdout) ;
	scanf("%d",&n) ;
	if(n<=120) ans = n / 6 ;
	else if(n > 1200) {
		ans = 220 + (n-1200)/3 - 3 ;
	}else {
		ans = 20 + int(double(n-120)/5.4) - 1;
 	}
 	cout << ans << endl ;
	return 0 ;
}

T4

思路

算出来一天少多少,然后与30比较

代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std ;
char s ;
double n ;
int m ;
int main () {
	freopen("fat.in","r",stdin) ;
	freopen("fat.out","w",stdout) ;
	cin >> s ;
	cin >> n >> m ;
	n = n * 2 ;
	int rest = s - 'A' + 2 ;
	rest -- ;
	if(m*rest >= 30) {
		double ans = (n - m*rest)/2.0 ;
		cout << ans <<endl ;
		puts("FLAG") ;
	}else {
		puts("-233333") ;
	}
	return 0 ;	
}

T5

思路

找出得到工资和因变丑而需要多添的钱之差的最大值然后,,,,

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std ;
int f , n , k , rest , tot = 0 ,maxx ;
int abs(int x) {
	return x > 0 ? x : -x ;
}
struct dy{
	int x , y ;
}a[10000] ;
int main () {
	freopen("facelift.in","r",stdin) ;
	freopen("facelift.out","w",stdout) ;
	scanf("%d%d%d",&f,&n,&k) ;
	rest = abs(f)*k ;
	for(int i = 1 ; i <= n ; i ++) {
		scanf("%d%d",&a[i].x,&a[i].y) ;
		maxx = max(a[i].x-a[i].y*k,maxx) ;
	}
	if(maxx){
		if(rest % maxx == 0) {
			cout << rest / maxx << endl ;
		}else {
			cout << rest / maxx + 1 <<endl ;
		}
	}
	else {
		puts("-666666") ;
	}
	return 0 ;
}

T6

思路

爆搜

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std ;
int a[12][12] ;
char s[12][12] ;
int vis[12][12] , h[12][12] , l[12][12] , small[10] , xie[12][12] ;
void dfs(int x,int y) {
	if(a[x][y]) {
		if(x == 9 && y == 9) {
			for(int i = 1 ; i <= 9 ; i ++) {
				for(int j = 1 ; j <= 9 ; j ++) {
					cout << a[i][j] << " " ;  
				}puts("") ;
			}
			return ;
		}
		if(y == 9) dfs(x+1,1) ;
		else dfs(x,y+1) ;
	}else {
		for(int i = 1 ; i <= 9 ; i ++) {
			if(!h[x][i] && !l[y][i] && !xie[(x-1)/3*3+(y-1)/3+1][i]) {
				a[x][y] = i ;
				h[x][i] = 1 ;
				l[y][i] = 1 ;
				xie[(x-1)/3*3+(y-1)/3+1][i] = 1 ;
				if(x == 9 && y == 9) {
					for(int i = 1 ; i <= 9 ; i ++) {
						for(int j = 1 ; j <= 9 ; j ++ ) {
							cout << a[i][j] << " " ;
						}puts("") ;
					}
					return ;
				}
				if(y == 9) dfs(x+1,1) ;
				else dfs(x,y+1) ;
				a[x][y] = 0 ;
				h[x][i] = 0 ;
				l[y][i] = 0 ;
				xie[(x-1)/3*3+(y-1)/3+1][i]=0;
			}
		}
	}
	
}

int main () {
	freopen("table.in","r",stdin) ;
	freopen("table.out","w",stdout) ;
	for(int i = 1 ; i <= 9 ; i ++) {
		for(int j = 1 ; j <= 9 ; j ++) {
			cin >> s[i][j] ;
		}
	}
	for(int i = 1 ; i <= 9 ; i ++) {
		for(int j = 1 ; j <= 9 ; j ++ ) {
			a[i][j] = s[i][j] - '0';			
		}
	}
	for(int i = 1 ; i <= 9 ; i ++) {
		for(int j = 1 ; j <= 9 ; j ++) {
			if(a[i][j] > 0) {
				h[i][a[i][j]] = 1 ;
				l[j][a[i][j]] = 1 ;
				xie[(i-1)/3*3+(j-1)/3+1][a[i][j]] = 1 ;
			}
		}
	}
	dfs(1,1) ;
	return 0 ;
}

该下去打球了8QAQ