无算法,判断(s - (a + b)) % 2是否为零,若零,表示在s步内还能走向其他的地方并且回来
否则,都是No
#include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <algorithm> #include <string> #include <map> #include <vector> #include <set> using namespace std; const int MAXN = 1e6 + 10; const int INF = 0x3f3f3f3f; int main(void) { //freopen ("A.in", "r", stdin); long long a, b, s; while (~scanf ("%I64d%I64d%I64d", &a, &b, &s)) { if (a < 0) a = -a; if (b < 0) b = -b; if (a + b <= s) { int x = s - (a + b); if (x % 2 == 0) puts ("Yes"); else puts ("No"); } else puts ("No"); } return 0; }
B. Drazil and His Happy Friends
无算法,标记和更新happy的人就行了
少写一个&!,导致runtime error
#include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <algorithm> #include <string> #include <map> #include <vector> #include <set> using namespace std; const int MAXN = 1e6 + 10; const int INF = 0x3f3f3f3f; int a[110], b[110]; int main(void) { //freopen ("B.in", "r", stdin); int n, m; scanf ("%d%d", &n, &m); memset (a, 0, sizeof (a)); memset (b, 0, sizeof (b)); int x, y, tmp, cnt = 0; scanf ("%d", &x); for (int i=0; i<x; ++i) { scanf ("%d", &tmp); a[tmp] = 1; cnt++; } scanf ("%d", &y); for (int i=0; i<y; ++i) { scanf ("%d", &tmp); b[tmp] = 1; cnt++; } for (int i=0; i<=m*n*2; ++i) { if (a[i%n] || b[i%m]) { if (!a[i%n]) cnt++; if (!b[i%m]) cnt++; a[i%n] = b[i%m] = 1; } } (cnt == n + m) ? puts ("Yes") : puts ("No"); return 0; }
无算法,数学问题貌似就是对单个数字分解质因数,替换,然后sort排序就行了
#include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <algorithm> #include <string> #include <map> #include <vector> #include <set> using namespace std; const int MAXN = 1e6 + 10; const int INF = 0x3f3f3f3f; string res[10] = {"", "", "2", "3", "322", "5", "53", "7", "7222", "7332"}; int main(void) { //freopen ("C.in", "r", stdin); int n; while (cin >> n) { string s, ans; cin >> s; for (int i=0; i<s.size(); ++i) { ans += res[s[i] - '0']; } sort (ans.begin (), ans.end ()); reverse (ans.begin (), ans.end ()); cout << ans << endl; } return 0; }