A
#include <cstdio>
int t, n, num;
int main() {
scanf("%d", &t);
while (t--) {
scanf("%d", &n); int odd = 0;
for (int i = 1; i <= n; i++) {
scanf("%d", &num);
if (num & 1) odd++;
}
if (odd == 0 || (!(odd & 1) && odd == n)) printf("NO\n");
else printf("YES\n");
}
return 0;
}
B
#include <cstdio>
int t, n;
int main() {
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
int ans = n;
while (n >= 10) {
int t = n / 10;
ans += t;
n += t - t * 10;
}
printf("%d\n", ans);
}
return 0;
}
C
有mp记录每个位置是否被访问过。 如果访问过代表走了个环。 求出走的路径长度。
#include <cstdio>
#include <algorithm>
#include <map>
#define mk(x,y) make_pair(x,y)
using namespace std;
const int N = 2e5 + 5;
char s[N];
int t, n;
int main() {
scanf("%d", &t);
while (t--) {
scanf("%d%s", &n, s + 1);
map<pair<int, int>, int> mp;
mp[mk(0, 0)] = 1; //作为起点
int x = 0, y = 0, l, r, minv = 1e9;
for (int i = 1; i <= n; i++) {
if (s[i] == 'L') x--;
else if (s[i] == 'R') x++;
else if (s[i] == 'U') y++;
else if (s[i] == 'D') y--;
pair<int, int> t = mk(x, y);
if (mp.count(t)) {
//取个最小值
int d = i - mp[t];
if (d < minv) {
minv = d; r = i, l = mp[t];
}
}
mp[mk(x, y)] = i + 1; //及时更新位置
}
if (minv == 1e9) printf("-1\n");
else printf("%d %d\n", l, r);
}
return 0;
}
A
#include <cstdio>
int t, n, num;
int main() {
s
阅读背景：
Codeforces Round #617 (Div. 3) A Array with Odd Sum B Food Buying C Yet Another Walking Robot
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