琐碎的区间
时间限制: 4 Sec 内存限制: 256 MB提交: 131 解决: 26
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题目描述
给出一个长度为 n 的整数序列 A[1..n]，有三种操作：
1 l r x : 把[l, r]区间的每个数都加上 x
2 l r : 把[l, r] 区间每个 A[i]变为sqrt(a[i])的整数部分
3 l r : 求[l, r] 区间所有数的和
其中 l 和 r 和 x 都代表一个整数
1 l r x : 把[l, r]区间的每个数都加上 x
2 l r : 把[l, r] 区间每个 A[i]变为sqrt(a[i])的整数部分
3 l r : 求[l, r] 区间所有数的和
其中 l 和 r 和 x 都代表一个整数
输入
第一行一个 T，表示数据组数。
对于每组数据
Line1：两个数 n m，表示整数序列长度和操作数
Line2：n 个数，表示 A[1..n]
Line3…Line3+m-1：每行一个询问，对于第三种询问，请输出答案。
对于每一种询问，先给出操作的编号，再给出相应的操作，编号与题目描述对应。
数据约定：
1<=T<=5
n,m <= 100000
1<= A[i], x<=100000
对于每组数据
Line1：两个数 n m，表示整数序列长度和操作数
Line2：n 个数，表示 A[1..n]
Line3…Line3+m-1：每行一个询问，对于第三种询问，请输出答案。
对于每一种询问，先给出操作的编号，再给出相应的操作，编号与题目描述对应。
数据约定：
1<=T<=5
n,m <= 100000
1<= A[i], x<=100000
输出
对于第三种询问，输出答案。每个答案占一行。
样例输入
1
5 5
1 2 3 4 5
1 3 5 2
2 1 4
3 2 4
2 3 5
3 1 5
样例输出
5
6
提示
来源
#include <bits/stdc++.h>
#define ll long long
const int maxn=1e5+10;
const int N=2e5+10;
using namespace std;
#define ls rt<<1
#define rs rt<<1|1
int n,m,k,t,a[maxn];
ll tag[maxn<<2],ma[maxn<<2],mi[maxn<<2],sum[maxn<<2];
const int BufferSize=1<<16;
char buffer[BufferSize],*head,*tail;
inline char Getchar() {
if(head==tail) {
int l=fread(buffer,1,BufferSize,stdin);
tail=(head=buffer)+l;
}
return *head++;
}
inline int read() {
int x=0,f=1;char c=Getchar();
for(;!isdigit(c);c=Getchar()) if(c=='-') f=-1;
for(;isdigit(c);c=Getchar()) x=x*10+c-'0';
return x*f;
}
void pup(int l,int r,int rt)
{
int mid=l+r>>1;
sum[rt]=sum[ls]+sum[rs];
ma[rt]=max(ma[ls],ma[rs]);
mi[rt]=min(mi[ls],mi[rs]);
tag[rt]=0;
}
void pdw(int l,int r,int rt)
{
int mid=l+r>>1;
sum[ls]+=tag[rt]*(mid-l+1);
ma[ls]+=tag[rt];
mi[ls]+=tag[rt];
tag[ls]+=tag[rt];
sum[rs]+=tag[rt]*(r-mid);
ma[rs]+=tag[rt];
mi[rs]+=tag[rt];
tag[rs]+=tag[rt];
tag[rt]=0;
}
void build(int l,int r,int rt)
{
if(l==r)
{
tag[rt]=0;
ma[rt]=mi[rt]=sum[rt]=a[l];
return;
}
int mid=l+r>>1;
build(l,mid,ls);
build(mid+1,r,rs);
pup(l,r,rt);
}
void upd(int L,int R,ll v,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
sum[rt]+=v*(r-l+1);
ma[rt]+=v;
mi[rt]+=v;
tag[rt]+=v;
return;
}
int mid=l+r>>1;
if(tag[rt])pdw(l,r,rt);
if(L<=mid)upd(L,R,v,l,mid,ls);
if(R>mid)upd(L,R,v,mid+1,r,rs);
pup(l,r,rt);
}
void qsqrt(int L,int R,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
if(ma[rt]==mi[rt])
{
tag[rt]-=ma[rt];
ma[rt]=sqrt(ma[rt]);
tag[rt]+=ma[rt];
mi[rt]=ma[rt];
sum[rt]=(r-l+1)*ma[rt];
return;
}
else if(ma[rt]==mi[rt]+1)
{
if((ll)sqrt(ma[rt])==(ll)sqrt(mi[rt])+1)
{
tag[rt]-=ma[rt];
sum[rt]-=(r-l+1)*(ma[rt]-(ll)sqrt(ma[rt]));
ma[rt]=sqrt(ma[rt]);
tag[rt]+=ma[rt];
mi[rt]=ma[rt]-1;
return;
}
}
}
int mid=l+r>>1;
if(tag[rt])pdw(l,r,rt);
if(L<=mid)qsqrt(L,R,l,mid,ls);
if(R>mid)qsqrt(L,R,mid+1,r,rs);
pup(l,r,rt);
}
ll gao(int L,int R,int l,int r,int rt)
{
if(L<=l&&r<=R)return sum[rt];
int mid=l+r>>1;
if(tag[rt])pdw(l,r,rt);
ll ret=0;
if(L<=mid)ret+=gao(L,R,l,mid,ls);
if(R>mid)ret+=gao(L,R,mid+1,r,rs);
return ret;
}
int main()
{
t=read();
while(t--)
{
n=read();m=read();
for(int i=1;i<=n;i++)
a[i]=read();
build(1,n,1);
while(m--)
{
int b,c,d,e;
b=read(),c=read(),d=read();
if(b==1)
{
e=read();
upd(c,d,e,1,n,1);
}
else if(b==2)
{
qsqrt(c,d,1,n,1);
}
else printf("%lld\n",gao(c,d,1,n,1));
}
}
return 0;
}PS超时的代码：
#include <bits/stdc++.h>
using namespace std;
const int BufferSize=1<<16;
char buffer[BufferSize],*head,*tail;
inline char Getchar() {
if(head==tail) {
int l=fread(buffer,1,BufferSize,stdin);
tail=(head=buffer)+l;
}
return *head++;
}
inline int read() {
int x=0,f=1;char c=Getchar();
for(;!isdigit(c);c=Getchar()) if(c=='-') f=-1;
for(;isdigit(c);c=Getchar()) x=x*10+c-'0';
return x*f;
}
#define LL long long
#define L(root) ((root) << 1)
#define R(root) (((root) << 1) | 1)
const int MAXN = 1e5 + 5;
int numbers[MAXN];
//LL delay[MAXN * 4], sum[MAXN * 4], mx[MAXN * 4], mn[MAXN * 4];
struct Node {
int left, right;
LL delay;
LL sum;
LL mx, mn;
int mid()
{
return left + ((right - left) >> 1);
}
} tree[MAXN * 4];
void pushUp(int root)
{
tree[root].sum = tree[L(root)].sum + tree[R(root)].sum;
tree[root].mx = max(tree[L(root)].mx, tree[R(root)].mx);
tree[root].mn = min(tree[L(root)].mn, tree[R(root)].mn);
tree[root].delay = 0;
}
void pushDown(int root, int l, int r)
{
LL mid = (r + l) >> 1;
tree[L(root)].delay += tree[root].delay;
tree[R(root)].delay += tree[root].delay;
tree[L(root)].sum += tree[root].delay * (mid - l + 1);
tree[R(root)].sum += tree[root].delay * (r - mid);
tree[L(root)].mx += tree[root].delay;
tree[R(root)].mx += tree[root].delay;
tree[L(root)].mn += tree[root].delay;
tree[R(root)].mn += tree[root].delay;
tree[root].delay = 0;
}
void build(int root, int left, int right)
{
tree[root].left = left;
tree[root].right = right;
if (left == right) {
tree[root].delay = 0;
tree[root].sum = numbers[left];
tree[root].mx = numbers[left];
tree[root].mn = numbers[left];
return;
}
int mid = tree[root].mid();
build(L(root), left, mid);
build(R(root), mid + 1, right);
pushUp(root);
}
LL query(int root, int left, int right)
{
if (tree[root].left == left && tree[root].right == right) {
return tree[root].sum;
}
if (tree[root].delay) pushDown(root, tree[root].left, tree[root].right);
int mid = tree[root].mid();
if (right <= mid) {
return query(L(root), left, right);
} else if (left > mid) {
return query(R(root), left, right);
} else {
return query(L(root), left, mid) + query(R(root), mid + 1, right);
}
}
void update(int root, int left, int right, LL add)
{
if (tree[root].left == left && tree[root].right == right) {
tree[root].delay += add;
tree[root].sum += add * (right - left + 1);
tree[root].mx += add;
tree[root].mn += add;
return;
}
if (tree[root].delay) pushDown(root, tree[root].left, tree[root].right);
int mid = tree[root].mid();
if (right <= mid) {
update(L(root), left, right, add);
} else if (left > mid) {
update(R(root), left, right, add);
} else {
update(L(root), left, mid, add);
update(R(root), mid + 1, right, add);
}
pushUp(root);
}
void sq(int root, int left, int right)
{
if (tree[root].left == left && tree[root].right == right) {
LL mx = sqrt(tree[root].mx);
LL mn = sqrt(tree[root].mn);
if (tree[root].mx == tree[root].mn) {
tree[root].delay -= (tree[root].mx - mx);
tree[root].sum = mx * (right - left + 1);
tree[root].mx = mx;
tree[root].mn = mn;
return;
// }
} else if ((tree[root].mx == tree[root].mn + 1) && (mx == mn + 1)) {
tree[root].delay -= (tree[root].mx - mx);
tree[root].sum -= (tree[root].mx - mx) * (right - left + 1);
tree[root].mx = mx;
tree[root].mn = mn;
}
}
if (tree[root].delay) pushDown(root, tree[root].left, tree[root].right);
int mid = tree[root].mid();
if (right <= mid) {
sq(L(root), left, right);
} else if (left > mid) {
sq(R(root), left, right);
} else {
sq(L(root), left, mid);
sq(R(root), mid + 1, right);
}
pushUp(root);
}
int main()
{
int t;
int n, m;
int i;
int op, l, r, x;
// scanf("%d", &t);
t = read();
while (t--) {
// scanf("%d%d", &n, &m);
n = read(), m = read();
for (i = 1; i <= n; ++i) {
// scanf("%d", &numbers[i]);
numbers[i] = read();
}
build(1, 1, n);
for (i = 0; i < m; ++i) {
// scanf("%d", &op);
op = read();
if (op == 1) {
// scanf("%d%d%d", &l, &r, &x);
l = read(), r = read(), x = read();
update(1, l, r, x);
//printf("debug op = 1\n");
} else if (op == 2) {
// scanf("%d%d", &l, &r);
l = read(), r = read();
sq(1, l, r);
//printf("debug op = 2\n");
} else {
// scanf("%d%d", &l, &r);
l = read(), r = read();
printf("%lld\n", query(1, l, r));
//printf("debug op = 3\n");
}
}
}
return 0;
}
int main2()
{
int t;
int n, m;
int i;
int op, l, r, x;
scanf("%d", &t);
// t = read();
while (t--) {
scanf("%d%d", &n, &m);
// n = read(), m = read();
for (i = 1; i <= n; ++i) {
scanf("%d", &numbers[i]);
// numbers[i] = read();
}
build(1, 1, n);
for (i = 0; i < m; ++i) {
scanf("%d", &op);
// op = read();
if (op == 1) {
scanf("%d%d%d", &l, &r, &x);
// l = read(), r = read(), x = read();
update(1, l, r, x);
//printf("debug op = 1\n");
} else if (op == 2) {
scanf("%d%d", &l, &r);
// l = read(), r = read();
sq(1, l, r);
//printf("debug op = 2\n");
} else {
scanf("%d%d", &l, &r);
// l = read(), r = read();
printf("%lld\n", query(1, l, r));
//printf("debug op = 3\n");
}
}
}
return 0;
}
