Codeforces Round #267 (Div. 2)
A:签到题,直接for一遍
B:取异或就是不同的数,然后bitcount一下判断即可
C:dp,dp[i]表示到i的最大值,然后对取与不取当前位置进行转移即可,要先把前缀和预处理出来
D:先利用map,把字符串hash掉,然后建图,现场在做的时候是直接记忆化搜索,不过这样处理不了环的情况,果断fst了,后来换了下姿势,先求强连通进行缩点,缩点完就是一个dag了,然后记忆化搜索即可
代码:
A:
#include int main() { int n, p, q; int ans = 0; scanf("%d", &n); while (n--) { scanf("%d%d", &p, &q); if (q - p >= 2) ans++; } printf("%d\n", ans); return 0;}
B:
#include #include int n, m, k;int bitcount(int x) { int ans = 0; while (x) { ans += (x&1); x >>= 1; } return ans;}const int N = 1005;int x[N];int main() { scanf("%d%d%d", &n, &m, &k); for (int i = 0; i <= m; i++) scanf("%d", &x[i]); int ans = 0; for (int i = 0; i < m; i++) { if (bitcount(x[i]^x[m]) <= k) ans++; } printf("%d\n", ans); return 0;}
C:
#include #include #include using namespace std;const int N = 5005;const long long INF = 0x3f3f3f3f3f3f3f;long long dp[N][N];int n, m, k;long long num[N], pre[N];int main() { scanf("%d%d%d", &n, &m, &k); for (int i = 1; i <= n; i++) { scanf("%lld", &num[i]); pre[i] = pre[i - 1] + num[i]; } for (int i = 0; i <= n; i++) for (int j = 0; j <= k; j++) dp[i][j] = -INF; dp[0][0] = 0; for (int i = 1; i <= n; i++) { for (int j = 0; j <= k; j++) { if (j && i >= m) dp[i][j] = max(dp[i][j], dp[i - m][j - 1] + pre[i] - pre[i - m]); dp[i][j] = max(dp[i][j], dp[i - 1][j]); } } printf("%lld\n", dp[n][k]); return 0;}
D:
#include #include #include #include #include #include #include #include