45 条题解
-
2
Sky1231 (sky1231) LV 10 @ 2020-10-20 20:40:51
#include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; namespace dts { struct st_node { int l,r,mid,sum,lazy; //sum:maxsum,lazy:maxsumlazy }; int n,m,a[100000]; st_node st[(1<<19)+2]; #define lc(now) ((now)<<1) #define rc(now) ((now)<<1|1) void st_push_down(int now) { st[lc(now)].sum+=st[now].lazy; st[rc(now)].sum+=st[now].lazy; st[lc(now)].lazy+=st[now].lazy; st[rc(now)].lazy+=st[now].lazy; st[now].lazy=0; } void st_build(int now,int l,int r) { st[now].l=l,st[now].r=r; st[now].lazy=0; if (l<r) { st[now].mid=(l+r)>>1; st_build(lc(now),l,st[now].mid); st_build(rc(now),st[now].mid+1,r); st[now].sum=max(st[lc(now)].sum,st[rc(now)].sum); } else st[now].sum=a[st[now].l-1]; } void st_update(int now,int l,int r,int val) { if (st[now].l==l&&r==st[now].r) st[now].sum+=val,st[now].lazy+=val; else { st_push_down(now); if (r<=st[now].mid) st_update(lc(now),l,r,val); else if (st[now].mid+1<=l) st_update(rc(now),l,r,val); else st_update(lc(now),l,st[now].mid,val),st_update(rc(now),st[now].mid+1,r,val); st[now].sum=max(st[lc(now)].sum,st[rc(now)].sum); } } int st_ask(int now,int l,int r) { if (st[now].l==l&&r==st[now].r) return st[now].sum; else { st_push_down(now); if (r<=st[now].mid) return st_ask(lc(now),l,r); else if (st[now].mid+1<=l) return st_ask(rc(now),l,r); else return max(st_ask(lc(now),l,st[now].mid),st_ask(rc(now),st[now].mid+1,r)); } } void main() { scanf("%d",&n); for (int i=0;i<n;i++) scanf("%d",&a[i]); st_build(1,1,n); scanf("%d",&m); for (int i=1;i<=m;i++) { int K,l,r; scanf("%d%d%d",&K,&l,&r); if (K==1) { int val; scanf("%d",&val); st_update(1,l,r,val); } else if (K==2) printf("%d\n",st_ask(1,l,r)); } } } int main() { dts::main(); } -
2
线段树区间修改+区间查询
```cpp
#include<cstdio>
using namespace std;int n,m,x,y,z;
struct data
{
int l,r,x,tag;
} tr[300005];int readln()
{
int x=0,f=1;
char ch=getchar();
while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
while ('0'<=ch&&ch<='9') x=x*10+ch-48,ch=getchar();
return x*f;
}int max(int x,int y){return x>y?x:y;}
void build(int l,int r,int rt)
{
tr[rt].l=l;tr[rt].r=r;
if (l==r) {tr[rt].x=readln();return;}
int mid=(l+r)>>1;
build(l,mid,rt<<1);build(mid+1,r,rt<<1|1);
tr[rt].x=max(tr[rt<<1].x,tr[rt<<1|1].x);
}void pushdown(int rt)
{
tr[rt<<1].tag+=tr[rt].tag;
tr[rt<<1|1].tag+=tr[rt].tag;
tr[rt<<1].x+=tr[rt].tag;
tr[rt<<1|1].x+=tr[rt].tag;
tr[rt].tag=0;
}void updata(int l,int r,int add,int rt)
{
int ll=tr[rt].l,rr=tr[rt].r;
if (ll==l&&rr==r) {tr[rt].tag+=add;tr[rt].x+=add;return;}
if (tr[rt].tag) pushdown(rt);
int mid=(ll+rr)>>1;
if (r<=mid) updata(l,r,add,rt<<1);
else if (mid<l) updata(l,r,add,rt<<1|1);
else {updata(l,mid,add,rt<<1);updata(mid+1,r,add,rt<<1|1);}
tr[rt].x=max(tr[rt<<1].x,tr[rt<<1|1].x);
}int query(int l,int r,int rt)
{
int ll=tr[rt].l,rr=tr[rt].r;
if (ll==l&&rr==r) return tr[rt].x;
if (tr[rt].tag) pushdown(rt);
int mid=(ll+rr)>>1;
if (r<=mid) return query(l,r,rt<<1);
else if (mid<l) return query(l,r,rt<<1|1);
else return max(query(l,mid,rt<<1),query(mid+1,r,rt<<1|1));
}int main()
{
n=readln();
build(1,n,1);
m=readln();
for (int i=1;i<=m;i++)
{
x=readln();
if (x==1){
x=readln();y=readln();z=readln();
updata(x,y,z,1);
}
else {
x=readln();y=readln();
printf("%d\n",query(x,y,1));
}
}
}
``` -
0
好基础的一道线段树啊……
#include<iostream> #include<cstdio> #include<cstring> using namespace std; struct tree { int l,r; long long maxn,lb; }a[10000005]; int n,m; int s[1000005]; void build(int d,int l,int r) { a[d].l=l; a[d].r=r; a[d].lb=0; if(l==r) { a[d].maxn=s[l]; return; } int mid=(l+r)/2; build(2*d,l,mid); build(2*d+1,mid+1,r); a[d].maxn=max(a[2*d].maxn,a[2*d+1].maxn); return; } inline void xf(int d) { if(a[d].lb!=0) { a[2*d].maxn+=a[d].lb; a[2*d+1].maxn+=a[d].lb; a[2*d].lb+=a[d].lb; a[2*d+1].lb+=a[d].lb; a[d].lb=0; } } void gai(int d,int x,int y,long long s) { if(a[d].l>=x&&a[d].r<=y) { a[d].maxn+=s; a[d].lb+=s; return; } xf(d); int mid=(a[d].l+a[d].r)/2; if(x<=mid) { gai(2*d,x,y,s); } if(y>mid) { gai(2*d+1,x,y,s); } a[d].maxn=max(a[2*d].maxn,a[2*d+1].maxn); return; } long long ask(int d,int x,int y) { if(a[d].l>=x&&a[d].r<=y) { return a[d].maxn; } xf(d); int mid=(a[d].l+a[d].r)/2; long long ans=-1e18; if(x<=mid) { ans=max(ans,ask(2*d,x,y)); } if(y>mid) { ans=max(ans,ask(2*d+1,x,y)); } return ans; } int main() { scanf("%d",&n); for(int i=1;i<=n;++i) { scanf("%d",&s[i]); } build(1,1,n); scanf("%d",&m); int x,y,op; long long s; for(int i=1;i<=m;++i) { scanf("%d",&op); if(op==1) { scanf("%d%d%lld",&x,&y,&s); gai(1,x,y,s); } else { scanf("%d%d",&x,&y); printf("%lld\n",ask(1,x,y)); } } return 0; } /* 5 1 2 3 4 5 3 2 1 4 1 1 3 3 2 3 5 */ -
0
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;const int N = 100000 + 5;
int n, m, a[N], addv[N << 2], maxv[N << 2];
inline void maintain(int rt) { maxv[rt] = max(maxv[rt << 1], maxv[rt << 1 | 1]); }
inline void pushdown(int rt) {
addv[rt << 1] += addv[rt];
addv[rt << 1 | 1] += addv[rt];
maxv[rt << 1] += addv[rt];
maxv[rt << 1 | 1] += addv[rt];
addv[rt] = 0;
}void build(int L, int R, int rt) {
if (L == R) {
maxv[rt] = a[L];
return;
}
int mid = (L + R) >> 1;
build(L, mid, rt << 1);
build(mid + 1, R, rt << 1 | 1);
maintain(rt);
}void update(int l, int r, int v, int L, int R, int rt) {
if (l <= L && r >= R) {
addv[rt] += v;
maxv[rt] += v;
return;
}
pushdown(rt);
int mid = (L + R) >> 1;
if (l <= mid) update(l, r, v, L, mid, rt << 1);
if (r > mid) update(l, r, v, mid + 1, R, rt << 1 | 1);
maintain(rt);
}int query(int l, int r, int L, int R, int rt) {
if (l <= L && r >= R) return maxv[rt];
pushdown(rt);
int mid = (L + R) >> 1;
if (r <= mid) return query(l, r, L, mid, rt << 1);
else if (l > mid) return query(l, r, mid + 1, R, rt << 1 | 1);
else return max(query(l, r, L, mid, rt << 1), query(l, r, mid + 1, R, rt << 1 | 1));
maintain(rt);
}int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
build(1, n, 1);
scanf("%d", &m);
for (int i = 1; i <= m; i++) {
int k, l, r, c; scanf("%d", &k);
if (k == 1) {
scanf("%d%d%d", &l, &r, &c);
update(l, r, c, 1, n, 1);
} else {
scanf("%d%d", &l, &r);
printf("%d\n", query(l, r, 1, n, 1));
}
}
return 0;
} -
0
1697749107 LV 8 @ 2016-11-17 20:44:52
在更新的时候一边pushup 一边pushdown
#include<cstdio>
#include<iostream>
#define L(u) (u<<1)
#define R(u) (u<<1|1)
using namespace std;
const int N=100005;
struct xx{
int l,r,val,Max;
}T[N<<2];
int n,t,a[N];
int op,L,R,C;
void Pushup(int u){T[u].Max=max(T[L(u)].Max,T[R(u)].Max);}
inline void Pushdown(int u)
{
if(!T[u].val)return;
T[L(u)].val+=T[u].val;
T[R(u)].val+=T[u].val;
T[L(u)].Max+=T[u].val;
T[R(u)].Max+=T[u].val;
T[u].val=0;
}
void Build(int u,int l,int r)
{
T[u].l=l,T[u].r=r;
if(l==r)
{
T[u].Max=a[l];
return ;
}
int mid=(l+r)>>1;
Build(L(u),l,mid),Build(R(u),mid+1,r);
Pushup(u);
}
inline void Update(int u,int l,int r,int c)
{
if(T[u].l==l&&T[u].r==r)
{
T[u].val+=c;
T[u].Max+=c;
return;
}
Pushdown(u);
int mid=(T[u].l+T[u].r)>>1;
if(l>mid)
Update(R(u),l,r,c);
else if(r<=mid)
Update(L(u),l,r,c);
else
Update(L(u),l,mid,c),Update(R(u),mid+1,r,c);
Pushup(u);
}
inline int Query(int u,int l,int r)
{
if(T[u].l==l&&T[u].r==r)
return T[u].Max;
Pushdown(u);
int mid=(T[u].l+T[u].r)>>1;
if(l>mid)
return Query(R(u),l,r);
else if(r<=mid)
return Query(L(u),l,r);
else
return max(Query(L(u),l,mid),Query(R(u),mid+1,r));
Pushup(u);
}
int main()
{
cin>>n;
for(int i=1;i<=n;i++)scanf("%d",a+i);
Build(1,1,n);
cin>>t;
while(t--)
{
scanf("%d",&op);
if(op==1)
{
scanf("%d%d%d",&L,&R,&C);
Update(1,L,R,C);
}
else{
scanf("%d%d",&L,&R);
printf("%d\n",Query(1,L,R));
}
}
return 0;
} -
0
#include<cstdio>
#include<iostream>
#define L(u) (u<<1)
#define R(u) (u<<1|1)
using namespace std;
const int N=100005;int n,t,a[N],op,L,R,C;struct xx{int l,r,val,Max;}T[N<<2];
inline void Pushdown(int u){if(!T[u].val)return;T[L(u)].val+=T[u].val,T[R(u)].val+=T[u].val,T[L(u)].Max+=T[u].val,T[R(u)].Max+=T[u].val;T[u].val=0;}
void Build(int u,int l,int r){T[u].l=l,T[u].r=r;if(l==r){T[u].Max=a[l];return;}int mid=(l+r)>>1;Build(L(u),l,mid),Build(R(u),mid+1,r);T[u].Max=max(T[L(u)].Max,T[R(u)].Max);}
inline void Update(int u,int l,int r,int c){if(T[u].l==l&&T[u].r==r){T[u].val+=c,T[u].Max+=c;return;}Pushdown(u); int mid=(T[u].l+T[u].r)>>1;if(l>mid)Update(R(u),l,r,c);else if(r<=mid)Update(L(u),l,r,c);else Update(L(u),l,mid,c),Update(R(u),mid+1,r,c);T[u].Max=max(T[L(u)].Max,T[R(u)].Max);}
inline int Query(int u,int l,int r){if(T[u].l==l&&T[u].r==r)return T[u].Max;Pushdown(u);int mid=(T[u].l+T[u].r)>>1;if(l>mid)return Query(R(u),l,r);else if(r<=mid)return Query(L(u),l,r);else return max(Query(L(u),l,mid),Query(R(u),mid+1,r));T[u].Max=max(T[L(u)].Max,T[R(u)].Max);}
int main(){cin>>n;for(int i=1;i<=n;i++)scanf("%d",a+i);Build(1,1,n);cin>>t;while(t--){scanf("%d",&op);if(op==1)scanf("%d%d%d",&L,&R,&C),Update(1,L,R,C);else scanf("%d%d",&L,&R),printf("%d\n",Query(1,L,R));}return 0;} -
0
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0
1048922798 LV 8 @ 2016-10-17 12:50:36
线段树区间更新+区间最大值。。线段树很不熟。。最坑的是写读入优化的时候偷懒直接没有负数的判断，我以为所有的数都是正的5555555555555555（~~o(>_<)o ~~） 然后就A了。。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define L(u) (u<<1)
#define R(u) (u<<1|1)
#define ll long long
using namespace std;
const int M=1000005;
struct Node{
int maxn,val,add;
int l,r;
}node[M<<2];
int A[M],m,n,k,p,l,flag,a,b,c;
inline ll read()
{
ll x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
void Pushup(int u){
node[u].maxn=max(node[L(u)].maxn,node[R(u)].maxn);
}
void Pushdown(int u){
node[L(u)].add+=node[u].add;
node[R(u)].add+=node[u].add;
node[L(u)].maxn+=node[u].add;
node[R(u)].maxn+=node[u].add;
node[u].add=0;
}
void Build(int u,int left,int right){
node[u].l=left;node[u].r=right;
if(node[u].l==node[u].r){
node[u].maxn=A[left];
return;
}
int mid=(left+right)>>1;
Build(L(u),left,mid);Build(R(u),mid+1,right);
Pushup(u);
}
void Update(int u,int left,int right,int v){
if(node[u].l==left&&node[u].r==right){
node[u].add+=v;
node[u].maxn+=v;
return;
}
if(node[u].add) Pushdown(u);
int mid=(node[u].l+node[u].r)>>1;
if(right<=mid) Update(L(u),left,right,v);
else if(left>=mid+1) Update(R(u),left,right,v);
else {Update(L(u),left,mid,v);Update(R(u),mid+1,right,v);}
Pushup(u);
}
int Query(int u,int left,int right){
if(node[u].l==left&&node[u].r==right){
return node[u].maxn;
}
if(node[u].add) Pushdown(u);
int mid=(node[u].l+node[u].r)>>1;
if(right<=mid) return Query(L(u),left,right);
else if(left>=mid+1) return Query(R(u),left,right);
else return max(Query(L(u),left,mid),Query(R(u),mid+1,right));
Pushup(u);
}
int main()
{
n=read();
for (int i=1;i<=n;i++)
A[i]=read();
Build(1,1,n);
m=read();
for (int i=1;i<=m;i++){
k=read();
if(k==2){a=read(),b=read();printf("%d\n",Query(1,a,b));}
if(k==1) {a=read(),b=read();c=read();Update(1,a,b,c);}
}
return 0;
} -
0
P1659河蟹王国
Accepted记录信息
评测状态 Accepted
题目 P1659 河蟹王国
递交时间 2015-07-31 12:32:48
代码语言 C++
评测机 VijosEx
消耗时间 2790 ms
消耗内存 10128 KiB
评测时间 2015-07-31 12:32:53评测结果
编译成功测试数据 #0: Accepted, time = 187 ms, mem = 3124 KiB, score = 10
测试数据 #1: Accepted, time = 328 ms, mem = 7088 KiB, score = 10
测试数据 #2: Accepted, time = 312 ms, mem = 10128 KiB, score = 10
测试数据 #3: Accepted, time = 165 ms, mem = 5688 KiB, score = 10
测试数据 #4: Accepted, time = 375 ms, mem = 8756 KiB, score = 10
测试数据 #5: Accepted, time = 250 ms, mem = 4544 KiB, score = 10
测试数据 #6: Accepted, time = 306 ms, mem = 6064 KiB, score = 10
测试数据 #7: Accepted, time = 281 ms, mem = 5832 KiB, score = 10
测试数据 #8: Accepted, time = 211 ms, mem = 2396 KiB, score = 10
测试数据 #9: Accepted, time = 375 ms, mem = 6280 KiB, score = 10
Accepted, time = 2790 ms, mem = 10128 KiB, score = 100
代码
#include <iostream>
#include <stdio.h>using namespace std;
struct Node
{
int l , r;
Node * left , * right;
int value;
int tag;
bool have;
};int max( int a , int b )
{
if( a > b )
return a;
return b;
}int merge( int a , int b )
{
return max( a , b );
}inline void pushdown( Node * cur )
{
if( cur -> l == cur -> r )
if( cur -> have )
{
cur -> have = 0;
cur -> value += cur -> tag;
cur -> tag = 0;
return;
}
if( cur -> have )
{
cur -> have = 0;
cur -> left -> tag += cur -> tag;
cur -> right -> tag += cur -> tag;
cur -> left -> have = 1;
cur -> right -> have = 1;
cur -> tag = 0;
cur -> value = merge( cur -> left -> value + cur -> left -> tag , cur -> right -> value + cur -> right -> tag );
}
return;
}inline void update( Node * cur )
{
cur -> value = merge( cur -> left -> value + cur -> left -> tag , cur -> right -> value + cur -> right -> tag );
return;
}int find( int l , int r , Node * cur )
{
//cout << l << " " << r << " " << cur -> l << " " << cur -> r << endl;
pushdown( cur );
if( cur -> l == l && cur -> r == r )
return cur -> value;
int mid = ( cur -> l + cur -> r ) / 2;
if( l > mid )
return find( l , r , cur -> right );
else if( mid >= r )
return find( l , r , cur -> left );
return merge( find( l , mid , cur -> left ) , find( mid + 1 , r , cur -> right ) );
}void modify( int l , int r , int v , Node * cur )
{
if( cur -> l == l && cur -> r == r )
{
cur -> have = 1;
cur -> tag += v;
return;
}
int mid = ( cur -> l + cur -> r ) / 2;
if( l > mid )
modify( l , r , v , cur -> right );
else if( mid >= r )
modify( l , r , v , cur -> left );
else
{
modify( l , mid , v , cur -> left );
modify( mid + 1 , r , v , cur -> right );
}
update( cur );
}int a[100000 + 2];
Node * build( int l , int r )
{
Node * cur = new Node();
cur -> l = l;
cur -> r = r;
cur -> tag = 0;
cur -> have = 0;
if( l == r )
{
cur -> value = a[l];
return cur;
}
int mid = ( l + r ) / 2;
cur -> left = build( l , mid );
cur -> right = build( mid + 1 , r );
cur -> value = merge( cur -> left -> value , cur -> right -> value );
return cur;
}Node * root;
int n , m;
int s;
int l , r , v;
int i;int main()
{
scanf( "%d" , &n );
for( i = 1 ; i <= n ; i++ )
scanf( "%d", &a[i] );
root = build( 1 , n );
scanf( "%d" , &m );
for( i = 0 ; i < m ; i++ )
{
scanf( "%d" , &s );
if( s == 2 )
{
scanf( "%d %d" , &l , &r );
printf( "%d\n" , find( l , r , root ) );
}
else
{
scanf( "%d %d %d" , &l , &r , &v );
modify( l , r , v , root );
}
}
return 0;
} -
0
lyk248289469 LV 9 @ 2013-11-26 14:28:32
虽然我还有负的这句话看了很多遍，但我自己写的时候初始化还是初始了0- -
-
0
貌似说，输出完全没问题啊！！！！WA了N次后，改成正常输出格式就对了！？？？？
编译成功
测试数据 #0: Accepted, time = 125 ms, mem = 6856 KiB, score = 10
测试数据 #1: Accepted, time = 125 ms, mem = 6856 KiB, score = 10
测试数据 #2: Accepted, time = 78 ms, mem = 6856 KiB, score = 10
测试数据 #3: Accepted, time = 31 ms, mem = 6856 KiB, score = 10
测试数据 #4: Accepted, time = 171 ms, mem = 6856 KiB, score = 10
测试数据 #5: Accepted, time = 140 ms, mem = 6856 KiB, score = 10
测试数据 #6: Accepted, time = 171 ms, mem = 6856 KiB, score = 10
测试数据 #7: Accepted, time = 140 ms, mem = 6856 KiB, score = 10
测试数据 #8: Accepted, time = 140 ms, mem = 6856 KiB, score = 10
测试数据 #9: Accepted, time = 187 ms, mem = 6856 KiB, score = 10
Accepted, time = 1308 ms, mem = 6856 KiB, score = 100#include <cstdio>
#include <algorithm>using namespace std;
#define MAXN 100001
int a[MAXN];
int n;struct saver{
int l,r;
int maxn;
int c;
} tree[MAXN*4];void make_tree(int l,int r,int k){
tree[k].l=l;
tree[k].r=r;
tree[k].c=0;
if (l==r){
tree[k].maxn=a[l];
return ;
}
int mid=(l+r)>>1;
make_tree(l,mid,k<<1);
make_tree(mid+1,r,k<<1^1);
tree[k].maxn=max(tree[k<<1].maxn,tree[k<<1^1].maxn);
}void Increase(int l,int r,int c,int k){
if (tree[k].l==l&&tree[k].r==r){
tree[k].c+=c;
return ;
}
int mid=(tree[k].l+tree[k].r)>>1;
if (r<=mid){ Increase(l,r,c,k<<1);} else
if (l>mid){ Increase(l,r,c,k<<1^1);} else
{
Increase(l,mid,c,k<<1);
Increase(mid+1,r,c,k<<1^1);
}
tree[k].maxn=max(tree[k<<1].maxn+tree[k<<1].c,tree[k<<1^1].maxn+tree[k<<1^1].c);
}int Search(int l,int r,int k){
if (l==tree[k].l&&r==tree[k].r){
return tree[k].c+tree[k].maxn;
}
int mid=(tree[k].l+tree[k].r)>>1;
if (r<=mid) return Search(l,r,k<<1)+tree[k].c;
if (l>mid) return Search(l,r,k<<1^1)+tree[k].c;
return max(Search(l,mid,k<<1),Search(mid+1,r,k<<1^1))+tree[k].c;
}int main(){
scanf("%d",&n);
for (int i=0;i++<n;) scanf("%d",&a[i]);
make_tree(1,n,1);
int m;
scanf("%d",&m);
while (m--){
int k;
scanf("%d",&k);
if (k==1){
int l,r,c;
scanf("%d%d%d",&l,&r,&c);
Increase(l,r,c,1);
}
else {
int l,r;
scanf("%d%d",&l,&r);
printf("%d\n",Search(l,r,1));
}
}
return 0;
} -
0
好河蟹的提交次数和通过人数-_-|||
Flag Accepted
题号 P1659
类型(?) 高级数据结构
通过 44人
提交 444次
通过率 10%
难度 2终于在下课铃响的时候过了……
编译通过...
├ 测试数据 01：答案正确... 9ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 25ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 25ms
├ 测试数据 08：答案正确... 25ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：84ms关于输入问题
都用read就ok了。
关于输出问题
我是每次write(find(l,r),' ');
在程序最后writeln;
ps:c是不是负数有关系吗？ -
0
JackDavid127 LV 10 @ 2009-10-08 12:44:50
第39个AC！！！！！！！！！
线段树的烦题
终于入门了！
ps：输出不能换行啊！
-
0
xiongnanbin LV 10 @ 2009-10-07 08:15:51
只能用wirte才能AC啊，真是晕死啊~
Sky1231 (sky1231)
LV 10
@ 2020-10-20 20:40:51
