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 ;
}
