版权声明：吸猫大法、 https://blog.csdn.net/sasuke__/article/details/82557184
A
题意:找出一段最长的区间, 是的这个区间内的字符串各不相同
思路:滑动窗口
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
// #include <unistd.h>
#include <unordered_map>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define fi first
#define se second
#define mst(a, b) memset(a, b, sizeof a)
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
const int MOD = 1e9 + 7;
char st[qq];
int ct[qq];
int main() {
#ifdef ONLINE_JUDGE
#else
freopen("in.txt", "r", stdin);
#endif
scanf("%s", st);
int n = strlen(st);
int maxn = 0, l = 0;
for (int i = 0; i < n; ++i) {
if (ct[st[i]] == 0) {
ct[st[i]]++;
} else {
while (ct[st[l]]--) {
l++;
if (ct[st[i]] == 0) break;
}
ct[st[i]]++;
}
maxn = max(maxn, i - l + 1);
}
printf("%d\n", maxn);
return 0;
}B
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
// #include <unistd.h>
#include <unordered_map>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define fi first
#define se second
#define mst(a, b) memset(a, b, sizeof a)
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
const int qq = 1e3 + 10;
const int INF = 1e9 + 10;
const int MOD = 1e9 + 7;
int num[qq][qq];
int n;
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
void dfs(int x, int y) {
num[x][y] = 0;
for (int i = 0; i < 4; ++i) {
int a = x + dx[i];
int b = y + dy[i];
if (a < 0 || b < 0 || a >= n || b >= n) continue;
if (num[a][b] == 0) continue;
dfs(a, b);
}
}
int main() {
#ifdef ONLINE_JUDGE
#else
freopen("in.txt", "r", stdin);
#endif
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
scanf("%d", &num[i][j]);
}
}
int Bcnt = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (num[i][j] == 1) {
dfs(i, j);
Bcnt++;
}
}
}
printf("%d\n", Bcnt);
return 0;
}C
题意: 求有多少种形成一个ip地址
思路: 实际是说在n - 1个空中选择三个点, 最终要能形成一个ip地址, 注意ip地址不带前缀0, 直接二进制枚举
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
// #include <unistd.h>
#include <unordered_map>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define fi first
#define se second
#define mst(a, b) memset(a, b, sizeof a)
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
const int MOD = 1e9 + 7;
char st[qq];
int getOne(int x) {
int cnt = 0;
while (x > 0) {
cnt++;
x = x & (x - 1);
}
return cnt;
}
bool check(int x, int y) {
int cnt = 0;
if (x == 0) cnt = 1;
while (x > 0) {
cnt++;
x /= 10;
}
if (cnt != y) return false;
return true;
}
int main() {
#ifdef ONLINE_JUDGE
#else
freopen("in.txt", "r", stdin);
#endif
scanf("%s", st);
int ans = 0;
int n = strlen(st) - 1;
// printf("%d\n", n);
for (int i = 0; i < (1 << n); ++i) {
int num = getOne(i);
if (num != 3) continue;
int pos[3], cnt = 0;
for (int j = 0; j < n; ++j) {
if (i & (1 << j)) {
pos[cnt++] = j + 1;
}
}
int a, b, c, d; a = b = c = d = 0;
for (int j = 0; j < pos[0]; ++j) {
a = a * 10 + st[j] - '0';
}
if (!check(a, pos[0])) continue;
for (int j = pos[0]; j < pos[1]; ++j) {
b = b * 10 + st[j] - '0';
}
if (!check(b, pos[1] - pos[0])) continue;
for (int j = pos[1]; j < pos[2]; ++j) {
c = c * 10 + st[j] - '0';
}
if (!check(c, pos[2] - pos[1])) continue;
for (int j = pos[2]; j < n + 1; ++j) {
d = d * 10 + st[j] - '0';
}
if (!check(d, (n + 1) - pos[2])) continue;
// printf("%d %d %d %d\n", a, b, c, d);
int flag = (1 << 8);
if (a < flag && b < flag && c < flag && d < flag) {
ans++;
}
}
printf("%d\n", ans);
return 0;
}D
题意: 判断是不是unicode编码的结果
思路: 模拟即可
这题的答案只有0和1, 从而可以推出你至少可以拿50%的分数, 假设总共100组数据, 答案为0是x组, 那么答案为y是 100 - x, 可知 你能拿的分数 = max(x, 100 - x), 只有当x = 50的时候取到最小值 50
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
// #include <unistd.h>
#include <unordered_map>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define fi first
#define se second
#define mst(a, b) memset(a, b, sizeof a)
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
const int qq = 1e5 + 10;
const int INF = 1e9 + 10;
const int MOD = 1e9 + 7;
int num[qq], n;
// LL bits[4];
bool check() {
if (n == 1) {
if (num[0] & (1 << 7)) return true;
return false;
}
for (int i = 1; i < n; ++i) {
if (!(num[i] & (1 << 7))) return false;
if (num[i] & (1 << 6)) return false;
}
for (int i = 0; i < n; ++i) {
if (!(num[0] & (1 << (7 - i)))) return false;
}
if (num[0] & (1 << (8 - n - 1))) return false;
return true;
}
int main() {
#ifdef ONLINE_JUDGE
#else
freopen("in.txt", "r", stdin);
#endif
// init();
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", num + i);
}
bool f = true;
for (int i = 0; i < n && f; ++i) {
int p = 0;
if (!(num[i] & (1 << 7))) {
p = 1;
} else {
int start = 7;
while (start >= 0 && (num[i] & (1 << start))) start--, p++;
if (p < 2) f = false;
if (p > 4) f = false;
if (i + p > n) f = false;
for (int j = i + 1; j < i + p; ++j) {
if (!(num[j] & (1 << 7))) f = false;
if (num[j] & (1 << 6)) f = false;
}
}
i += (p - 1);
}
int ans = f ? 1 : 0;
printf("%d\n", ans);
return 0;
}E
题意: 求抖音红人
思路: 这题需要前置技能强连通分量. 那么进行缩点形成DAG图以后, 统计出度为0的点的个数, 可以想想如果存在2个及以上的点, 很明显这两个点是互不能到达的. 实际可以看成缩点之后形成了一条链, 才有满足条件的答案. 注意还需要判断联通快的个数
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <bitset>
// #include <unistd.h>
// #include <unordered_map>
using namespace std;
#define LL long long
#define pb push_back
#define mk make_pair
#define pill pair<int, int>
#define fi first
#define se second
#define mst(a, b) memset(a, b, sizeof a)
#define lson (rt << 1)
#define rson ((rt << 1) | 1)
const int qq = 2e5 + 10;
const int INF = 1e9 + 10;
const int MOD = 1e9 + 7;
int n, m, Bcnt, top, Index;
int fa[qq], dfn[qq], low[qq], Stack[qq], belong[qq], num[qq], deg[qq];
bool instack[qq];
vector<int> vt[qq];
map<pill, bool> mp;
int find(int x) {
return fa[x] == -1 ? x : fa[x] = find(fa[x]);
}
void tarjan(int u) {
low[u] = dfn[u] = ++Index;
instack[u] = true;
Stack[top++] = u;
int sz = (int)vt[u].size();
for (int i = 0; i < sz; ++i) {
int v = vt[u][i];
if (!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
} else if (instack[v]) {
low[u] = min(low[u], dfn[v]);
}
}
int v;
if (low[u] == dfn[u]) {
++Bcnt;
do {
v = Stack[--top];
instack[v] = false;
belong[v] = Bcnt;
num[Bcnt]++;
} while (v != u);
}
}
void init() {
mst(fa, -1);
mst(num, 0);
mst(dfn, 0);
mst(deg, 0);
mst(instack, false);
mp.clear();
for (int i = 0; i <= n; ++i) {
vt[i].clear();
}
}
int main() {
#ifdef ONLINE_JUDGE
#else
freopen("in.txt", "r", stdin);
#endif
scanf("%d%d", &n, &m);
init();
for (int i = 0, a, b; i < m; ++i) {
scanf("%d%d", &a, &b);
if (a > n || b > n) continue;
if (mp.find(mk(a, b)) != mp.end()) continue;
mp[mk(a, b)] = true;
if (a == b) continue;
vt[a].pb(b);
int x = find(a), y = find(b);
if (x == y) continue;
fa[y] = x;
}
int p = 0;
for (int i = 1; i <= n; ++i) {
if (fa[i] == -1) p++;
}
if (p > 1) {
printf("0\n");
return 0;
}
Bcnt = top = Index = 0;
for (int i = 1; i <= n; ++i) {
if (!dfn[i]) {
tarjan(i);
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < (int)vt[i].size(); ++j) {
int v = vt[i][j];
if (belong[i] == belong[v]) continue;
deg[belong[i]]++;
}
}
int cnt = 0, id;
for (int i = 1; i <= Bcnt; ++i) {
if (deg[i] == 0) cnt++, id = i;
}
if (cnt == 1) {
printf("%d\n", num[id]);
} else {
printf("0\n");
}
return 0;
}