点击标题可传送
A-Transfer
输出c -= (a - b)
,如果(cle0)输出(0)
B - Uneven Numbers
给你一个(n),输出(1dots n)中位数为奇数的数字有几个
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int n;
int main() {
char t[200000];
scanf("%d", &n);
int cnt = 0;
for(int i = 1; i <= n; i++) {
sprintf(t, "%d", i);
if(strlen(t) & 1) cnt ++;
}
cout << cnt << '
';
return 0;
}
C - Build Stairs
给你(N)个方块,它们各自的高度为(H_i)。对于这些方块,你必须对它们做一次以下操作:
- 让方块的高度减(1)
- 什么也不做
将排序后的方块存到一个数组中,然后对排序后的和原先的作差,若差值大于(1)则输出No
,否则输出Yes
D - Gathering Children
一道模拟+思维的题目,经过推导可得到一些规律:
- 小孩最终会在
RL
这一组位置 - 当(r(L,overbrace {Rdots ,R}^r,L)+l(R,overbrace{L,dots ,L}^l,R))为偶数时,(r = l = frac{l+r}{2}).
- 当(r>l)时,如果(r)为偶数,则(r=lfloor frac{l+r}{2} floor ,l=lceil frac{l+r}{2} ceil);如果(r)为奇数,则(r=lceil frac{l+r}{2} ceil,l=lfloor frac{l+r}{2} floor).
- 当(l>r)时,如果(l)为奇数,则(l=lfloor frac{l+r}{2} floor,r=lceil frac{l+r}{2} ceil);如果(l)为偶数,则(l=lceil frac{l+r}{2} ceil,r=lfloor frac{l+r}{2} floor).
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1e5 + 3;
char s[N];
int a[N], l, r;
int main() {
scanf("%s", s);
int n = strlen(s);
for(int i = 0; i < n; i++) {
if(s[i] == 'R') r++;
else a[i] = r, r = 0;
}
for(int i = n - 1; i >= 0; i--) {
if(s[i] == 'L') l++;
else a[i] = l, l = 0;
}
// 经过上述两个for循环后,字符串中‘R’对应的是l的值,'L'对应的是r的值,顺序别搞错了
for(int i = 1; i < n; i++) {
if(s[i - 1] == 'R' && s[i] == 'L') {
int sum = a[i - 1] + a[i];
if(sum & 1) {
//当r > l时
if(a[i - 1] < a[i]) {
if(a[i] & 1) {
a[i - 1] = sum - sum / 2;
a[i] = sum / 2;
} else {
a[i - 1] = sum / 2;
a[i] = sum - sum / 2;
}
//当l > r时
} else if(a[i - 1] > a[i]){
if(a[i - 1] & 1) {
a[i - 1] = sum / 2;
a[i] = sum - sum / 2;
} else {
a[i - 1] = sum - sum / 2;
a[i] = sum / 2;
}
}
} else {
a[i - 1] = a[i] = sum >> 1;
}
}
}
for(int i = 0; i < n; i++) printf("%d ", a[i]); puts("");
return 0;
}