Codeforces Round #267 (Div. 2)_html/css_WEB-ITnose

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>

B:

#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>

D:

#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>