JZOJ 2022.02.11【提高A组】模拟

Solution

首先把 T2 给切了,T1 找半天规律找不到
然后打了个表算是暴力了
T3 也暴。。。
太暴了。。。
T4 直接啥也不会

T1

考虑一个 a 的答案
(c+b)(cb)=a2(c+b)(cb)=(a+1)(a1)c,b 组数
就是一般的分类讨论了,记一个数 a 的答案为 f(a)d(a)a 的约数个数
那么 2a,f(a)=d(a)/2
4a,f(a)=d(a/4)/2
其余则为 0
具体考虑 a2a21f 值即可

Code

#include <cstdio>
#define RE register
using namespace std;
typedef long long LL;
const int N = 1e7 + 5;
int x, y, d[N], _d[N], num[N], tot, pr[N / 10], vis[N], ans;
void Sieve()
{
d[1] = _d[1] = num[1] = 1;
for(RE int i = 2; i <= 1e7 + 3; i++)
{
if (!vis[i]) pr[++tot] = i, d[i] = 2, num[i] = 1, _d[i] = 3;
for(RE int j = 1, z; j <= tot && pr[j] * i <= 1e7 + 3; j++)
{
vis[z = i * pr[j]] = 1;
if (i % pr[j] == 0)
{
num[z] = num[i] + 1, d[z] = d[i] / num[z] * (num[z] + 1);
_d[z] = _d[i] / (num[i] * 2 + 1) * (num[z] * 2 + 1);
break;
}
else d[z] = d[i] * 2, num[z] = 1, _d[z] = _d[i] * 3;
}
}
}
int main()
{
Sieve(), scanf("%d%d", &x, &y);
for(RE int a = x; a <= y; a++)
if (a & 1) ans += (LL)_d[a] / 2 + (LL)d[(a + 1) / 2] * d[(a - 1) / 2] / 2;
else ans += (LL)d[a + 1] * d[a - 1] / 2 + _d[a / 2] / 2;
printf("%d\n", ans);
}

T2

一眼分位变成线段树赋值和异或操作

Code

#include <cstdio>
#define RE register
#define IN inline
using namespace std;
const int N = 5e4 + 5, LG = 9, SZ = N * 4, M = LG + 2;
int n, m, a[N];
struct SegmenTree{
#define ls (p << 1)
#define rs (ls | 1)
int sum[SZ][M], tg1[SZ][M], tg2[SZ][M];
IN void pushup(int p, int k){sum[p][k] = sum[ls][k] + sum[rs][k];}
void build(int p, int l, int r)
{
for(RE int j = 0; j <= LG; j++) tg1[p][j] = -1;
if (l == r)
{
for(RE int j = 0; j <= LG; j++) sum[p][j] = ((a[l] >> j) & 1);
return;
}
int mid = l + r >> 1; build(ls, l, mid), build(rs, mid + 1, r);
for(RE int j = 0; j <= LG; j++) pushup(p, j);
}
IN void push(int p, int k, int z, int l, int r)
{
if (z == 0 || z == 1) sum[p][k] = z * (r - l + 1), tg1[p][k] = z;
else{
sum[p][k] = r - l + 1 - sum[p][k], tg2[p][k] ^= 1;
if (tg1[p][k] != -1) tg1[p][k] ^= 1;
}
}
IN void pushdown(int p, int k, int l, int r)
{
int mid = l + r >> 1;
if (tg2[p][k]) push(ls, k, 2, l, mid), push(rs, k, 2, mid + 1, r), tg2[p][k] = 0;
if (tg1[p][k] != -1) push(ls, k, tg1[p][k], l, mid), push(rs, k, tg1[p][k], mid + 1, r), tg1[p][k] = -1;
}
void assign(int p, int l, int r, int k, int x, int y, int z)
{
if (x > r || y < l) return;
if (x <= l && r <= y) return push(p, k, z, l, r), void();
pushdown(p, k, l, r); int mid = l + r >> 1;
if (x <= mid) assign(ls, l, mid, k, x, y, z);
if (y > mid) assign(rs, mid + 1, r, k, x, y, z);
pushup(p, k);
}
void reverse(int p, int l, int r, int k, int x, int y)
{
if (x > r || y < l) return;
if (x <= l && r <= y) return push(p, k, 2, l, r), void();
pushdown(p, k, l, r); int mid = l + r >> 1;
if (x <= mid) reverse(ls, l, mid, k, x, y);
if (y > mid) reverse(rs, mid + 1, r, k, x, y);
pushup(p, k);
}
int Query(int p, int l, int r, int k, int x, int y)
{
if (x > r || y < l) return 0;
if (x <= l && r <= y) return sum[p][k];
pushdown(p, k, l, r); int mid = l + r >> 1, res = 0;
if (x <= mid) res = Query(ls, l, mid, k, x, y);
if (y > mid) res += Query(rs, mid + 1, r, k, x, y);
return res;
}
}T;
int main()
{
scanf("%d%d", &n, &m);
for(RE int i = 1; i <= n; i++) scanf("%d", &a[i]);
T.build(1, 1, n); char op[10];
for(RE int l, r, x; m; --m)
{
scanf("%s%d%d", op, &l, &r);
if (op[0] == 'a'){
scanf("%d", &x);
for(RE int j = 0; j <= LG; j++) if (!((x >> j) & 1)) T.assign(1, 1, n, j, l, r, 0);
}
else if (op[0] == 'o'){
scanf("%d", &x);
for(RE int j = 0; j <= LG; j++) if ((x >> j) & 1) T.assign(1, 1, n, j, l, r, 1);
}
else if (op[0] == 'x'){
scanf("%d", &x);
for(RE int j = 0; j <= LG; j++) if ((x >> j) & 1) T.reverse(1, 1, n, j, l, r);
}
else if (op[1] == 'x'){
int res = 0, z;
for(RE int j = 0; j <= LG; j++) z = T.Query(1, 1, n, j, l, r), res += ((z & 1) ? (1 << j) : 0);
printf("%d\n", res);
}
else{
int res = 0, z;
for(RE int j = 0; j <= LG; j++) z = T.Query(1, 1, n, j, l, r), res += z * (1 << j);
printf("%d\n", res);
}
}
}
posted @   leiyuanze  阅读(30)  评论(0编辑  收藏  举报
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