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

2019/08/18 练习赛

原文：https://www.cnblogs.com/lyt020321/p/11371467.html