T1 Permut
题意：
求\(1 - n\)的排列中逆序对数量为\(k\)的排列的个数
SOL：
排除法我们知道一定不是\(O(n!)\)的算法
考虑\(dp\)，现在已经有\(n-1\)的答案了，考虑新加入一个数产生多少新的逆序对
设\(dp[i][j]\)表示\(1 -i\)的排列有\(j\)个逆序对的数量，考虑新加入的数插在哪里会增加多少逆序对数量
有\[dp[i][j] = \sum\limits ^{min(i - 1, j)}_k dp[i - 1][j - k]\]
看起来有点奇怪，变一下形 \[dp[i][j] = \sum\limits^j_{max(0, j - i + 1} dp[i - 1][k]\]
复杂度\(O(n * k ^2)\)，\(O(跑不动)\)
考虑一个前缀和优化，类似“我要长高”和“\(Making the Grade\)”里的一样
由于\(k\)的上界随\(j\)的变化而变化，考虑在循环的时候累加\(dp[i - 1][j]\)到一个变量里，然后赋给\(dp[i][j]\)
注意这里\(k\)的下界在\(j - i + 1 \geq 0\)的时候会发生变化，记得把多加的减掉
#include<bits/stdc++.h>
#define N (1000 + 10)
using namespace std;
int T;
int n, k, sum, f[N][N];
const int mod = 10000;
int main() {
scanf("%d", &T);
f[1][0] = 1;
while (T--) {
scanf("%d%d", &n, &k);
if (f[n][k]) {printf("%d\n", f[n][k]); continue;}
for (register int i = 2; i <= n; ++i) {
sum = 0;
for (register int j = 0; j <= k; ++j) {
sum += f[i - 1][j] % mod;
f[i][j] = sum % mod;
if (j - i + 1 >= 0) sum -= f[i - 1][j - i + 1] % mod;
}
}
printf("%d\n", f[n][k]);
}
return 0;
}T2 Beautiful
题意：
定义一个数的值\(v_i\)为以它作为中位数的区间的最大长度，\(Q\)次查询询问区间\([l, r]\)内的最大\(v_i\)，原数大小比较时按值为第一关键字，下标为第二关键字
\(n \leq 2000, Q \leq 100000\)
SOL：
看到\(n\)很小，想想能不能怎么操作一下呢
考虑\(O(n)\)枚举每一个数，然后分别向左向右扩展并记录一个变量\(S\)，遇到比它大的就\(--S\)，否则\(++S\)，然后对于每个\(S\)的值记录一个最远位置，最后拼在一起取最大长度得到\(v_i\)
然后随便怎么做一下\(RMQ\)即可
#include<bits/stdc++.h>
#define N (100000 + 10)
using namespace std;
inline int read() {
int cnt = 0, f = 1; char c = getchar();
while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();}
return cnt * f;
}
int n, Q, l, r;
int S1[N], S2[N];
int a[N], b[N];
int t[N];
int A[N];
struct node {
int l, r;
int gmax;
#define l(p) tree[p].l
#define r(p) tree[p].r
#define gmax(p) tree[p].gmax
}tree[N << 2];
void pushup(int p) {
gmax(p) = max(gmax(p << 1), gmax(p << 1 | 1));
}
void build(int p, int l, int r) {
l(p) = l, r(p) = r;
if (l == r) {gmax(p) = A[l]; return;}
int mid = (l + r) >> 1;
build (p << 1, l, mid);
build (p << 1 | 1, mid + 1, r);
pushup(p);
}
long long query(int p, int l, int r) {
if (l <= l(p) && r >= r(p)) return gmax(p);
int mid = (l(p) + r(p)) >> 1;
long long ans = -1;
if (l <= mid) ans = max(ans, query(p << 1, l, r));
if (r > mid) ans = max(ans, query(p << 1 | 1, l, r));
return ans;
}
int main() {
n = read(); for (register int i = 1; i <= n; ++i) a[i] = read();
for (register int i = 1; i <= n; ++i) {
int tmp = 0;
memset(S1, 255, sizeof(S1));
memset(S2, 255, sizeof(S2));
S1[n] = S2[n] = 0;
for (register int j = i - 1; j >= 1; --j) {
if (a[j] > a[i]) ++tmp; if (a[j] <= a[i]) --tmp;
S1[tmp + n] = i - j;
}
tmp = 0;
for (register int j = i + 1; j <= n; ++j) {
if (a[j] >= a[i]) ++tmp; if (a[j] < a[i]) --tmp;
S2[tmp + n] = j - i;
}
for (register int j = 1 - i; j <= i - 1; ++j) if (S1[n + j] >= 0 && S2[n - j] >= 0) A[i] = max(A[i], S1[n + j] + 1 + S2[n - j]);
} Q = read();
// for (register int i = 1; i <= n; ++i) cout<<A[i]<<" ";return 0;
build (1, 1, n);
while (Q--) {
l = read(), r = read();
printf("%lld\n", query(1, l, r));
}
return 0;
}T3 Subset
题意：
维护一个集合\(A\)，支持插入删除查询\(a_i\&S=a_i\)的个数，\(a_i \in A\)，操作\(2e5\)，数字大小\(2^{16}\)
SOL：
奇妙的思路
由二进制去考虑，把数拆成两节，记录\(s[a][b]\)表示前八位是\(a\)，后八位是\(b\)的子集的个数
对于\(add\)和\(del\)定\(b\)枚举\(a\)更新
对于\(cnt\)操作定\(a\)枚举\(b\)更新
平衡了查询和修改之间的复杂度，时间复杂度优化为\(O(n * 2 ^ 8)\)
#include<bits/stdc++.h>
#define N ((1 << 8) + 5)
using namespace std;
int Q, x;
char ope[5];
int s[N][N];
inline int read() {
int cnt = 0, f = 1; char c = getchar();
while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();}
return cnt * f;
}
void add(int x) {
int a = x >> 8, b = x - (a << 8);
for (register int i = 0; i < (1 << 8); ++i) if ((i & a) == a) s[i][b]++;
}
void del(int x) {
int a = x >> 8, b = x - (a << 8);
for (register int i = 0; i < (1 << 8); ++i) if ((i & a) == a) s[i][b]--;
}
int query(int x) {
int ans = 0, a = x >> 8, b = x - (a << 8);
for (register int i = 0; i < (1 << 8); ++i) if ((i & b) == i) ans += s[a][i];
return ans;
}
int main() {
Q = read();
while (Q--) {
scanf("%s", ope + 1); x = read();
if (ope[1] == 'a') add(x);
if (ope[1] == 'd') del(x);
if (ope[1] == 'c') printf("%d\n", query(x));
}
return 0;
}原文地址：https://www.cnblogs.com/kma093/p/11621761.html