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 <cstdio>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;}</cstdio>
#include <cstdio>#include <cstring>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 <br> C: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>#include <algorithm>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 = 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;}</algorithm></cstring></cstdio>#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>#include <string>#include <iostream>#include <map>using namespace std;typedef long long ll;const int N = 1000005;vector<int> g[N], scc[N];int pre[N], lowlink[N], sccno[N], dfs_clock, scc_cnt;stack<int> S;typedef pair<ll ll> pii;int n, m, hn;map<string int> hash;string str[N];pii s[N];pii p[N];pii dp[N];int vis[N];vector<int> g2[N];void dfs_scc(int u) { pre[u] = lowlink[u] = ++dfs_clock; S.push(u); for (int i = 0; i = 'A' && str[i] > str[i]; get(str[i]); } scanf("%d", &m); string u, v; while (m--) { cin >> u >> v; int uu = get(u); int vv = get(v); g[uu].push_back(vv); } find_scc(hn); ll ans1 = 0, ans2 = 0; for (int i = 0; i <br> <br> <p></p> </int></string></ll></int></int></map></iostream></string></algorithm></stack></vector></cstring></cstdio>
