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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF551E GukiZ and GukiZiana

LuoGu: CF551E GukiZ and GukiZiana

CF: E. GukiZ and GukiZiana

暴力分块,块内排序。

查看代码
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#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
typedef long long LL;
const int N = 5e5 + 10, M = 710;
int n, m, st[M], id[N];
LL w[N], v[N], add[M];
int main ()
{
read(n), read(m);
int blk = sqrt(n);
for (int i = 1; i <= n; i++)
{
read(w[i]), v[i] = w[i];
id[i] = i / blk;
}
for (int i = 0; i <= id[n]; i++)
st[i] = i * blk;
st[0] = 1, st[id[n] + 1] = n + 1;
for (int i = 0; i <= id[n]; i++)
sort(v + st[i], v + st[i + 1]);
for (int op; m; m--)
{
read(op);
if (op == 1)
{
int l, r, x;
read(l), read(r), read(x);
if (id[l] == id[r])
{
for (int i = l; i <= r; i++)
w[i] += x;
for (int i = st[id[l]]; i < st[id[l] + 1]; i++)
v[i] = w[i];
sort(v + st[id[l]], v + st[id[l] + 1]);
continue;
}
for (int i = l; i < st[id[l] + 1]; i++)
w[i] += x;
for (int i = st[id[l]]; i < st[id[l] + 1]; i++)
v[i] = w[i];
sort(v + st[id[l]], v + st[id[l] + 1]);
for (int i = st[id[r]]; i <= r; i++)
w[i] += x;
for (int i = st[id[r]]; i < st[id[r] + 1]; i++)
v[i] = w[i];
sort(v + st[id[r]], v + st[id[r] + 1]);
for (int i = id[l] + 1; i < id[r]; i++)
add[i] += x;
}
else if (op == 2)
{
int y;
read(y);
int l = 0, r = 0;
for (int i = 0, t; !l && i <= id[n]; i++)
if ((t = lower_bound(v + st[i], v + st[i + 1], y - add[i]) - v) ^ st[i + 1] && (v[t] + add[i] == y))
for (int j = st[i]; !l && j < st[i + 1]; j++)
w[j] + add[i] == y && (l = j);
for (int i = id[n], t; !r && ~i; i--)
if ((t = lower_bound(v + st[i], v + st[i + 1], y - add[i]) - v) ^ st[i + 1] && (v[t] + add[i] == y))
for (int j = st[i + 1] - 1; !r && j >= st[i]; j--)
w[j] + add[i] == y && (r = j);
l ? (write(r - l), puts("")) : puts("-1");
}
}
return 0;
}