[Loj] 数列分块入门 1 - 9
数列分块入门 1
区间加 + 单点查询
#include <iostream>
#include <cstdio>
#include <cmath> using namespace std;
const int N = 5e4 + ; #define gc getchar() inline int read() {
int x = ; char c = gc;
while(c < '' || c > '') c = gc;
while(c >= '' && c <= '') x = x * + c - '', c = gc;
return x;
} int A[N], Add[N], bel[N];
int n, block, cnt; void Sec_G(int x, int y, int w) {
if(bel[x] == bel[y]) for(int i = x; i <= y; i ++) A[i] += w;
else {
for(int i = x; i <= bel[x] * block; i ++) A[i] += w;
for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] += w;
}
for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] += w;
} int main() {
n = read();
block = sqrt(n);
for(int i = ; i <= n; i ++) A[i] = read();
for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + ;
if(n % block) cnt = n / block + ;
else cnt = n / block;
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(!opt) Sec_G(l, r, c);
else cout << A[r] + Add[bel[r]] << endl;
}
return ;
}
数列分块入门 2
区间加法，询问区间内小于某个值 x 的元素个数
用B[]记录A[], B[]数组中为排好序的A[]的映射
那么每次可以对每一块进行二分查找
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm> using namespace std;
const int N = 5e4 + ; #define gc getchar() inline int read() {
int x = ; char c = gc;
while(c < '' || c > '') c = gc;
while(c >= '' && c <= '') x = x * + c - '', c = gc;
return x;
} int A[N], B[N], Add[N], bel[N];
int n, block, cnt; void Work_sort(int x) {
int l = (x - ) * block + , r = min(l + block - , n);
for(int i = l; i <= r; i ++) B[i] = A[i];
sort(B + l, B + r + );
} void Sec_G(int x, int y, int w) {
if(bel[x] == bel[y]) {
for(int i = x; i <= y; i ++) A[i] += w;
Work_sort(bel[x]);
}
else {
for(int i = x; i <= bel[x] * block; i ++) A[i] += w; Work_sort(bel[x]);
for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] += w; Work_sort(bel[y]);
}
for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] += w;
} inline int Calc(int x, int w) {
int l = (x - ) * block + , r = min(l + block - , n), ret = ;
while(l <= r) {
int mid = (l + r) >> ;
if(B[mid] + Add[x] < w) ret = mid, l = mid + ;
else r = mid - ;
}
return ret ? (ret - (x - ) * block) : ;
} inline int Sec_A(int x, int y, int w) {
int ret();
if(bel[x] == bel[y]) {
for(int i = x; i <= y; i ++) if(A[i] + Add[bel[x]] < w) ret ++;
return ret;
} else {
for(int i = x; i <= bel[x] * block; i ++) if(A[i] + Add[bel[x]] < w) ret ++;
for(int i = (bel[y] - ) * block + ; i <= y; i ++)
if(A[i] + Add[bel[y]] < w)
ret ++;
}
for(int i = bel[x] + ; i < bel[y]; i ++)
ret += Calc(i, w);
return ret;
} int main() {
n = read();
block = sqrt(n);
for(int i = ; i <= n; i ++) A[i] = read();
for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + ;
if(n % block) cnt = n / block + ;
else cnt = n / block;
for(int i = ; i <= cnt; i ++) Work_sort(i);
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(!opt) Sec_G(l, r, c);
else cout << Sec_A(l, r, c * c) << "\n";
}
return ;
}
数列分块入门 3
区间加法，询问区间内小于某个值 x 的前驱
与2类似，二分查找
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm> using namespace std;
const int N = 1e5 + ;
const int oo = ; #define gc getchar() inline int read() {
int x = ; char c = gc;
while(c < '' || c > '') c = gc;
while(c >= '' && c <= '') x = x * + c - '', c = gc;
return x;
} int A[N], B[N] = {-}, Add[N], bel[N];
int n, block, cnt; void Work_sort(int x) {
int l = (x - ) * block + , r = min(l + block - , n);
for(int i = l; i <= r; i ++) B[i] = A[i];
sort(B + l, B + r + );
} void Sec_G(int x, int y, int w) {
if(bel[x] == bel[y]) {
for(int i = x; i <= y; i ++) A[i] += w;
Work_sort(bel[x]);
}
else {
for(int i = x; i <= bel[x] * block; i ++) A[i] += w; Work_sort(bel[x]);
for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] += w; Work_sort(bel[y]);
}
for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] += w;
} inline int Calc(int x, int w) {
int l = (x - ) * block + , r = min(l + block - , n), ret = ;
while(l <= r) {
int mid = (l + r) >> ;
if(B[mid] + Add[x] < w) ret = mid, l = mid + ;
else r = mid - ;
}
return B[ret] + Add[x];
} inline int Sec_A(int x, int y, int w) {
int ret = -;
if(bel[x] == bel[y]) {
for(int i = x; i <= y; i ++) if(A[i] + Add[bel[x]] < w && A[i] + Add[bel[x]] > ret) ret = A[i] + Add[bel[x]];
return ret;
} else {
for(int i = x; i <= bel[x] * block; i ++)
if(A[i] + Add[bel[x]] < w && A[i] + Add[bel[x]] > ret) ret = A[i] + Add[bel[x]];
for(int i = (bel[y] - ) * block + ; i <= y; i ++)
if(A[i] + Add[bel[y]] < w && A[i] + Add[bel[y]] > ret) ret = A[i] + Add[bel[y]];
}
for(int i = bel[x] + ; i < bel[y]; i ++) {
int imp = Calc(i, w);
if(imp < w && imp > ret) ret = imp;
}
return ret;
} int main() {
n = read();
block = sqrt(n);
for(int i = ; i <= n; i ++) A[i] = read();
for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + ;
if(n % block) cnt = n / block + ;
else cnt = n / block;
for(int i = ; i <= cnt; i ++) Work_sort(i);
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(!opt) Sec_G(l, r, c);
else cout << Sec_A(l, r, c) << "\n";
}
return ;
}
数列分块入门 4
区间加法，区间求和
没什么好说的
#include <iostream>
#include <cstdio>
#include <cmath> using namespace std;
const int N = 5e4 + ; #define gc getchar() inline int read() {
int x = ; char c = gc;
while(c < '' || c > '') c = gc;
while(c >= '' && c <= '') x = x * + c - '', c = gc;
return x;
} #define LL long long LL A[N], Add[N], bel[N], W[N];
int n, block, cnt, Mod; void Sec_G(int x, int y, int w) {
if(bel[x] == bel[y]) for(int i = x; i <= y; i ++) A[i] += w, W[bel[x]] += w;
else {
for(int i = x; i <= bel[x] * block; i ++) A[i] += w, W[bel[x]] += w;
for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] += w, W[bel[y]] += w;
}
for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] += w, W[i] += w * block;
} inline int Sec_A(int x, int y) {
LL ret = ;
if(bel[x] == bel[y])
for(int i = x; i <= y; i ++)
ret += (A[i] + Add[bel[x]]) % Mod;
else {
for(int i = x; i <= bel[x] * block; i ++) ret += (A[i] + Add[bel[x]]) % Mod;
for(int i = (bel[y] - ) * block + ; i <= y; i ++) ret += (A[i] + Add[bel[y]]) % Mod;
}
for(int i = bel[x] + ; i < bel[y]; i ++) ret += W[i] % Mod;
return ret% Mod;
} int main() {
n = read();
block = sqrt(n);
for(int i = ; i <= n; i ++) A[i] = read();
for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + , W[bel[i]] += A[i];
if(n % block) cnt = n / block + ;
else cnt = n / block;
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read(); Mod = c + ;
if(!opt) Sec_G(l, r, c);
else cout << Sec_A(l, r) << endl;
}
return ;
}
数列分块入门 5
区间开方，区间求和
一个数(合理)开几次根后就是0/1了
因此，只需记录每块的最大值，如果最大值是0/1就没必要开根
#include <iostream>
#include <cstdio>
#include <cmath> using namespace std;
const int N = 5e4 + ; #define gc getchar() inline int read() {
int x = ; char c = gc;
while(c < '' || c > '') c = gc;
while(c >= '' && c <= '') x = x * + c - '', c = gc;
return x;
} #define LL long long LL A[N], Add[N], bel[N], W[N], Max[N];
int n, block, cnt, Mod; inline void Sec_G(int x, int y) {
if(bel[x] == bel[y]) {
if(!Max[bel[x]]) return ;
if(Max[bel[x]] == ) return ;
for(int i = x; i <= y; i ++) {
int C = A[i] - (int) sqrt(A[i]);
W[bel[x]] -= C; A[i] = (int) sqrt(A[i]);
}
LL Max_A = ;
for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) Max_A = max(Max_A, A[i]);
Max[bel[x]] = Max_A;
return ;
}
else {
if(Max[bel[x]] && Max[bel[x]] != ) {
for(int i = x; i <= bel[x] * block; i ++) {
int C = A[i] - (int) sqrt(A[i]);
W[bel[x]] -= C;
A[i] = (int) sqrt(A[i]);
}
LL Max_A = ;
for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) Max_A = max(Max_A, A[i]);
Max[bel[x]] = Max_A;
}
if(Max[bel[y]] && Max[bel[y]] != ) {
for(int i = (bel[y] - ) * block + ; i <= y; i ++) {
int C = A[i] - (int) sqrt(A[i]);
W[bel[y]] -= C;
A[i] = (int) sqrt(A[i]);
}
LL Max_A = ;
for(int i = (bel[y] - ) * block + ; i <= bel[y] * block; i ++) Max_A = max(Max_A, A[i]);
Max[bel[y]] = Max_A;
}
}
for(int i = bel[x] + ; i < bel[y]; i ++) {
if(!Max[i] || Max[i] == ) continue ;
LL Max_A = ;
for(int j = (i - ) * block + ; j <= i * block; j ++) {
int C = A[j] - (int) sqrt(A[j]);
W[i] -= C;
A[j] = (int) sqrt(A[j]);
Max_A = max(Max_A, A[j]);
}
Max[i] = Max_A;
}
} inline int Sec_A(int x, int y) {
LL ret = ;
if(bel[x] == bel[y] && Max[bel[x]]) for(int i = x; i <= y; i ++) ret += A[i];
else {
for(int i = x; i <= bel[x] * block && Max[bel[x]]; i ++) ret += A[i];
for(int i = (bel[y] - ) * block + ; i <= y && Max[bel[y]]; i ++) ret += A[i];
}
for(int i = bel[x] + ; i < bel[y]; i ++) ret += W[i];
return ret;
} int main() {
n = read();
block = sqrt(n);
for(int i = ; i <= n; i ++) A[i] = read();
for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + , W[bel[i]] += A[i], Max[bel[i]] = max(Max[bel[i]], A[i]);
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(!opt) Sec_G(l, r);
else cout << Sec_A(l, r) << "\n";
}
return ;
}
数列分块入门 6
单点插入，单点询问
数据随机，分块，对于每一块开动态数组，插入 + 查询比较容易实现
如果数据不随机，就有可能加到同一块中的数较多，影响效率
这样可以进行一定的插入操作之后重新分块
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector> using namespace std;
const int N = 1e5 + ; int A[N << ], n;
vector <int> Vec[];
int block, bel[N]; #define gc getchar() inline int read() {
int x = ; char c = gc;
while(c < '' || c > '') c = gc;
while(c >= '' && c <= '') x = x * + c - '', c = gc;
return x;
} inline void Ins(int x, int a) {
int now_size(), Whi;
for(int i = ; ; i ++) {
int Size = Vec[i].size();
now_size += Size;
if(now_size >= x) {
Whi = i;
x -= (now_size - Size);
break;
}
}
Vec[Whi].insert(Vec[Whi].begin() + x - , a);
} inline int Poi_A(int x) {
int Whi_, now_size();
for(int i = ; ; i ++) {
int Size = Vec[i].size();
now_size += Size;
if(now_size >= x) {
int iii = x - (now_size - Size);
return Vec[i][iii - ];
}
}
} int main() {
n = read();
block = sqrt(n);
for(int i = ; i <= n; i ++) A[i] = read();
for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + ;
for(int i = ; i <= n; i ++) Vec[bel[i]].push_back(A[i]);
int T = n; while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(!opt) Ins(l, r);
else cout << Poi_A(r) << endl;
}
return ;
}
数列分块入门 7
区间乘法，区间加法，单点询问
先乘后加，乘的时候相应的加法标记也要乘
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector> using namespace std;
const int N = 1e5 + ;
const int Mod = 1e4 + ; #define LL long long LL A[N], Mul[N], Add[N], bel[N];
int n, cnt, block; #define gc getchar() inline int read() {
int x = , f = ; char c = gc;
while(c < '' || c > '') {if(c == '-') f = -; c = gc;}
while(c >= '' && c <= '') x = x * + c - '', c = gc;
return x;
} inline void Sec_Add(int x, int y, int w) {
if(bel[x] == bel[y]) {
for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[x]] + Add[bel[x]]);
Add[bel[x]] = ; Mul[bel[x]] = ;
for(int i = x; i <= y; i ++) A[i] += w, A[i] %= Mod;
return ;
} else {
for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;
Add[bel[x]] = ; Mul[bel[x]] = ;
for(int i = x; i <= bel[x] * block; i ++) A[i] += w, A[i] %= Mod;
for(int i = (bel[y] - ) * block + ; i <= bel[y] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;
Add[bel[y]] = ; Mul[bel[y]] = ;
for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] += w, A[i] %= Mod;
}
for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] += w, Add[i] %= Mod;
} inline void Sec_Mul(int x, int y, int w) {
if(bel[x] == bel[y]) {
for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;
Add[bel[x]] = ; Mul[bel[x]] = ;
for(int i = x; i <= y; i ++) A[i] = (A[i] * w) % Mod;
return ;
} else {
for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;
Add[bel[x]] = ; Mul[bel[x]] = ;
for(int i = x; i <= bel[x] * block; i ++) A[i] = (A[i] * w) % Mod;
for(int i = (bel[y] - ) * block + ; i <= bel[y] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;
Add[bel[y]] = ; Mul[bel[y]] = ;
for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] = (A[i] * w) % Mod;
}
for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] = (Add[i] * w) % Mod, Mul[i] = (Mul[i] * w) % Mod;
} int main() {
n = read();
for(int i = ; i <= n; i ++) A[i] = read();
block = sqrt(n);
for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + , Mul[i] = ;
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(opt == ) Sec_Add(l, r, c);
else if(opt == ) Sec_Mul(l, r, c);
else cout << (A[r] * Mul[bel[r]] + Add[bel[r]]) % Mod << "\n";
}
return ;
}
数列分块入门 8
暴力
区间修改没有什么难度，这题难在区间查询比较奇怪，因为权值种类比较多，似乎没有什么好的维护方法。
模拟一些数据可以发现，询问后一整段都会被修改，几次询问后数列可能只剩下几段不同的区间了。
我们思考这样一个暴力，还是分块，维护每个分块是否只有一种权值，区间操作的时候，对于同权值的一个块就O(1)统计答案，否则暴力统计答案，并修改标记，不完整的块也暴力。
这样看似最差情况每次都会耗费O(n)的时间，但其实可以这样分析：
假设初始序列都是同一个值，那么查询是O(√n)，如果这时进行一个区间操作，它最多破坏首尾2个块的标记，所以只能使后面的询问至多多2个块的暴力时间，所以均摊每次操作复杂度还是O(√n)。
换句话说，要想让一个操作耗费O(n)的时间，要先花费√n个操作对数列进行修改。
初始序列不同值，经过类似分析后，就可以放心的暴力啦。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring> using namespace std;
const int N = 1e5 + ; int bel[N], A[N], bec[N];
int n;
int block; #define gc getchar() inline int read() {
int x = ; char c = gc;
while(c < '' || c > '') c = gc;
while(c >= '' && c <= '') x = x * + c - '', c = gc;
return x;
} int Sec_A(int x, int y, int c) {
int ret();
if(bel[x] == bel[y]) {
if(bec[bel[x]] == c) ret = y - x + ;
else if(bec[bel[x]] == -) for(int i = x; i <= y; i ++) if(A[i] == c) ret ++;
if(~ bec[bel[x]]) for(int i = (bel[x] - ) * block + ; i < x; i ++) A[i] = bec[bel[x]];
if(~ bec[bel[x]]) for(int i = y + ; i <= bel[x] * block; i ++) A[i] = bec[bel[x]];
for(int i = x; i <= y; i ++) A[i] = c;
bec[bel[x]] = -;
} else {
if(bec[bel[x]] == c) ret += bel[x] * block - x + ;
else if(bec[bel[x]] == -) for(int i = x; i <= bel[x] * block; i ++) if(A[i] == c) ret ++;
if(bec[bel[y]] == c) ret += y - ((bel[y] - ) * block);
else if(bec[bel[y]] == -) for(int i = (bel[y] - ) * block + ; i <= y; i ++) if(A[i] == c) ret ++;
for(int i = bel[x] + ; i < bel[y]; i ++) {
if(bec[i] == c) ret += block;
else if(bec[i] == -)
for(int j = (i - ) * block + ; j <= i * block; j ++)
if(A[j] == c) ret ++;
}
if(~ bec[bel[x]]) for(int i = (bel[x] - ) * block + ; i < x; i ++) A[i] = bec[bel[i]];
for(int i = x; i <= bel[x] * block; i ++) A[i] = c;
bec[bel[x]] = -;
for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] = c;
if(~ bec[bel[y]]) for(int i = y + ; i <= bel[y] * block; i ++) A[i] = bec[bel[i]];
bec[bel[y]] = -;
for(int i = bel[x] + ; i < bel[y]; i ++) bec[i] = c;
}
return ret;
} int main() {
n = read();
for(int i = ; i <= n; i ++) bec[i] = -;
for(int i = ; i <= n; i ++) A[i] = read();
block = sqrt(n);
for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + ;
int T = n;
while(T --) {
int l = read(), r = read(), c = read();
cout << Sec_A(l, r, c) << "\n";
}
return ;
}
数列分块入门 9
区间众数查询
陈立杰区间众数解题报告
#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define mod 10007
#define pi acos(-1)
#define inf 0x7fffffff
#define ll long long
using namespace std;
ll read() {
ll x=,f=;
char ch=getchar();
while(ch<''||ch>'') {
if(ch=='-')f=-;
ch=getchar();
}
while(ch>=''&&ch<='') {
x=x*+ch-'';
ch=getchar();
}
return x*f;
}
int n,blo,id;
int v[],bl[];
int f[][];
map<int,int>mp;
int val[],cnt[];
vector<int>ve[];
void pre(int x) {
memset(cnt,,sizeof(cnt));
int mx=,ans=;
for(int i=(x-)*blo+; i<=n; i++) {
cnt[v[i]]++;
int t=bl[i];
if(cnt[v[i]]>mx||(cnt[v[i]]==mx&&val[v[i]]<val[ans]))
ans=v[i],mx=cnt[v[i]];
f[x][t]=ans;
}
}
int query(int l,int r,int x) {
int t=upper_bound(ve[x].begin(),ve[x].end(),r)-lower_bound(ve[x].begin(),ve[x].end(),l);
return t;
}
int query(int a,int b) {
int ans,mx;
ans=f[bl[a]+][bl[b]-];
mx=query(a,b,ans);
for(int i=a; i<=min(bl[a]*blo,b); i++) {
int t=query(a,b,v[i]);
if(t>mx||(t==mx&&val[v[i]]<val[ans]))ans=v[i],mx=t;
}
if(bl[a]!=bl[b])
for(int i=(bl[b]-)*blo+; i<=b; i++) {
int t=query(a,b,v[i]);
if(t>mx||(t==mx&&val[v[i]]<val[ans]))ans=v[i],mx=t;
}
return ans;
}
int main() {
n=read();
blo=;
for(int i=; i<=n; i++) {
v[i]=read();
if(!mp[v[i]]) {
mp[v[i]]=++id;
val[id]=v[i];
}
v[i]=mp[v[i]];
ve[v[i]].push_back(i);
}
for(int i=; i<=n; i++)bl[i]=(i-)/blo+;
for(int i=; i<=bl[n]; i++)pre(i);
for(int i=; i<=n; i++) {
int a=read(),b=read();
if(a>b)swap(a,b);
printf("%d\n",val[query(a,b)]);
}
return ;
}
分块算法小结：
暴力算法
时间复杂度可以
空间允许
优美
[Loj] 数列分块入门 1 - 9的更多相关文章
- loj 数列分块入门 6 9(区间众数)
6 题意 给出一个长为\(n\)的数列,以及\(n\)个操作,操作涉及单点插入,单点询问,数据随机生成. 题解 参考:http://hzwer.com/8053.html 每个块内用一个\(vecto ...
- loj 数列分块入门 5 7 8
5 题意 给出一个长为\(n\)的数列,以及\(n\)个操作,操作涉及区间开方,区间求和. 思路 用\(tag\)记录这一块是否已全为\(1\). 除分块外,还可用 树状数组+并查集(链表) 或者 线 ...
- LOJ 数列分块入门系列
目录 1.区间加+单点查 每个块维护tag,散的暴力改. code: #include<bits/stdc++.h> using namespace std; const int maxn ...
- LOJ 6277：数列分块入门 1（分块入门）
#6277. 数列分块入门 1 内存限制:256 MiB时间限制:100 ms标准输入输出 题目类型:传统评测方式:文本比较 上传者: hzwer 提交提交记录统计讨论 3 测试数据 题目描述 给出一 ...
- LOJ #6285. 数列分块入门 9-分块(查询区间的最小众数)
#6285. 数列分块入门 9 内存限制:256 MiB时间限制:1500 ms标准输入输出 题目类型:传统评测方式:文本比较 上传者: hzwer 提交提交记录统计测试数据讨论 2 题目描述 给 ...
- LOJ #6284. 数列分块入门 8-分块(区间查询等于一个数c的元素，并将这个区间的所有元素改为c)
#6284. 数列分块入门 8 内存限制:256 MiB时间限制:500 ms标准输入输出 题目类型:传统评测方式:文本比较 上传者: hzwer 提交提交记录统计测试数据讨论 2 题目描述 给出 ...
- LOJ #6283. 数列分块入门 7-分块(区间乘法、区间加法、单点查询)
#6283. 数列分块入门 7 内存限制:256 MiB时间限制:500 ms标准输入输出 题目类型:传统评测方式:文本比较 上传者: hzwer 提交提交记录统计测试数据讨论 2 题目描述 给出 ...
- LOJ #6282. 数列分块入门 6-分块(单点插入、单点查询、数据随机生成)
#6282. 数列分块入门 6 内存限制:256 MiB时间限制:500 ms标准输入输出 题目类型:传统评测方式:文本比较 上传者: hzwer 提交提交记录统计测试数据讨论 1 题目描述 给出 ...
- LOJ #6281. 数列分块入门 5-分块(区间开方、区间求和)
#6281. 数列分块入门 5 内存限制:256 MiB时间限制:500 ms标准输入输出 题目类型:传统评测方式:文本比较 上传者: hzwer 提交提交记录统计测试数据讨论 5 题目描述 给出 ...
随机推荐
- vue—组件通信，ref
组件通信: 父组件传递子组件: 把需要的数据 传递给 子组件的数据,以数据绑定(v-bind)的形式,传递到子组件内部,供子组件使用,缩写(:) 动态传递: 第一步:在父组件中的子组件标签中进行动态的 ...
- 文件 open 方法
文件对象方法: 文件对象方法 执行操作 f.close() 关闭文件 f.read([size=-1]) 从文件读取size个字符,当未给定size或给定负值的时候, 读取剩余的所有字符,然后 ...
- print() 默认是打印完字符串会自动添加一个换行符
可以使用end=" " 参数告诉 print() 用空格代替换行 for i in range(1,10): ... print(i,end=' ') ... 1 2 3 4 5 ...
- javascript——HTML对象
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title> ...
- 用于Linq的去重 Distinct
/// <summary> /// 用于Linq的去重,扩展方法需要放到静态类中 /// </summary> ...
- 9. Java分支语句之if...else
if...else条件语句 一个if语句包含一个布尔表达式和一条或者多条语句. 语法运用有三种 //第一种 if(布尔表达式){ //如果布尔表达式为true将执行的语句 } //第二种 if(布尔表 ...
- 很low的四位验证码实现
<html> <head> <meta charset="utf-8"> </head> <body> <inpu ...
- 小程序 wxs时间戳转字符串
function formatDate(value) { //不能使用 new Date() var time = getDate(value); var year = time.getFullYea ...
- css3控制字体动态变换颜色
css3控制字体动态变换颜色 <!doctype html> <html> <head> <meta charset="utf-8"> ...
- 程序员和IT员不能错过的网站
前言本文收录一些值得收藏的开发相关网站. 目录一.搜索引擎二.在线课程三.在线练习四.在线工具箱五.在线编译器六.技术论坛或社区七.音乐放松一下 一.搜索引擎搜索引擎大家最熟悉不过,本没有必要列出,但 ...