蓝桥杯每日一题
(1)5369.棋盘
(2)4655.重新排序
(3)3745.牛的学术圈I
(4)1238.日志统计
以下是部分题目的代码
//棋盘(差分 | 异或运算)
const int N = 2010;
int n, m, a[N][N];
int main()
{
ios::sync_with_stdio(false);
cin >> n >> m;
while (m--)
{
int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
a[x1][y1] ^= 1;
a[x2 + 1][y1] ^= 1;
a[x1][y2 + 1] ^= 1;
a[x2 + 1][y2 + 1] ^= 1;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
{
a[i][j] ^= a[i - 1][j];
a[i][j] ^= a[i][j - 1];
a[i][j] ^= a[i - 1][j - 1];
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
cout << a[i][j];
cout << endl;
}
return 0;
}
//重新排序
typedef long long ll;
const ll N = 100010;
ll n, m, a[N], s[N], b[N], sum1, sum2;
int main()
{
ios::sync_with_stdio(false);
cin >> n;
for (ll i = 1; i <= n; i++)
{
cin >> a[i];
s[i] = a[i] + s[i - 1];
}
cin >> m;
while (m--)
{
ll l, r; cin >> l >> r;
b[l] += 1, b[r + 1] -= 1;//构建差分
sum1 += s[r] - s[l - 1];//重新排列前的总和
}
for (ll i = 1; i <= n; i++) b[i] += b[i - 1];//还原差分,得出每个元素被访问到的次数
//对两数组进行排序,令数组大的元素出现在访问次数多的位置
//进行排序后,索引越大,元素越大且访问次数越多
sort(a + 1, a + 1 + n), sort(b + 1, b + 1 + n);
for (ll i = 1; i <= n; i++) sum2 += a[i] * b[i];//重新排列后的总和
cout << sum2 - sum1 << endl;
return 0;
}
//日志统计
//数据相对复杂,看代码很容易懂,自己很难想出来,记得及时回顾
typedef pair<int, int> PII;
const int N = 100010;
PII logs[N];//记录时间和id
int cnt[N];//记录某个id的点赞数
bool st[N];//记录该id是否为热帖
int n, d, k;
int main()
{
ios::sync_with_stdio(false);
cin >> n >> d >> k;
for (int i = 1; i <= n; i++)
{
int ts, id; cin >> ts >> id;
logs[i] = { ts,id };
}
sort(logs + 1, logs + 1 + n);
for (int l = 1, r = 1; r <= n; r++)
{
int ts = logs[r].first, id = logs[r].second;
cnt[id]++;//给有记录的id的点赞数加一
//关键:根据时间差移动指针
while (logs[r].first - logs[l].first >= d)
{
cnt[logs[l].second]--;
l++;
}
if (cnt[id] >= k) st[id] = true;//标记为热帖
}
//遍历所有id
//这里不能遍历logs数组,因为有的id会重复,导致重复打印
for (int id = 1; id <= 100000; id++)
if (st[id]) cout << id << endl;
return 0;
}
//牛的学术圈I
//解法一(双指针)
const int N = 100010;
int n, m, a[N];
bool cmp(int x,int y)
{
return x > y;
}
int main()
{
ios::sync_with_stdio(false);
cin >> n >> m;
for (int i = 1; i <= n; i++) cin >> a[i];
sort(a + 1, a + 1 + n, cmp);//从大到小排序
int idx = 0;
//定义l和r,令位于[l,r]的元素相等
for (int l = 1, r = 1; r <= n;r++)
{
if (a[r] != a[l]) l = r;//更新区间
if (a[r] < r)
{
if (m == 0)
{
idx = r - 1;//记录索引
break;
}
for (int i = l; i <= n && i <= l + m - 1; i++) a[i]++;//增加引用次数
m = 0;//用光
r--;//抵消r++,在下个循环中重新判断当前元素是否>=索引
}
if (r == n) idx = n;//全部元素都>=索引
}
cout << idx << endl;
return 0;
}