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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#171173 | #7108. Couleur | xianggui# | AC ✓ | 1478ms | 34448kb | C++20 | 2.9kb | 2023-09-09 16:34:06 | 2023-09-09 16:34:54 |
Judging History
answer
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#define mid (l+r)/2
using namespace std;
const int N = 100010;
int n, m, cnt;
int a[N], b[N], T[N];
int sum[N<<5], L[N<<5], R[N<<5];
struct node {
int l,r;
long long ans;
};
set<node> S;
multiset<long long> ansS;
bool operator < (const node &a,const node &b){
return a.l<b.l||a.l==b.l&&a.r<b.r;
}
inline int build(int l, int r){
int rt = ++ cnt;
sum[rt] = 0;
if (l < r){
L[rt] = build(l, mid);
R[rt] = build(mid+1, r);
}
return rt;
}
inline int update(int pre, int l, int r, int x){
int rt = ++ cnt;
L[rt] = L[pre]; R[rt] = R[pre]; sum[rt] = sum[pre]+1;
if (l < r){
if (x <= mid) L[rt] = update(L[pre], l, mid, x);
else R[rt] = update(R[pre], mid+1, r, x);
}
return rt;
}
inline int query1(int u, int v, int l, int r, int val){
if (l >= r) return 0;
int x = sum[L[v]] - sum[L[u]];
if (val > mid) return x + query1(R[u], R[v], mid+1, r, val);
else return query1(L[u], L[v], l, mid, val);
}
inline int query2(int u, int v, int l, int r, int val){
if (l >= r) return sum[v] - sum[u];
int x = sum[L[v]] - sum[L[u]];
if (val > mid) return x + query2(R[u], R[v], mid+1, r, val);
else return query2(L[u], L[v], l, mid, val);
}
inline int getsum1(int x,int y,int z){ // < z
if (x>y) return 0;
return query1(T[x-1], T[y], 1, m, z);
}
inline int getsum2(int x,int y,int z){ // > z
if (x>y) return 0;
return (y-x+1)-query2(T[x-1], T[y], 1, m, z);
}
void solve(){
S.clear();ansS.clear();
scanf("%d",&n);m=0;cnt=0;
for (int i = 1; i <= n; i ++){
scanf("%d",&a[i]);
m = max(m,a[i]);
}
T[0] = build(1, m);
for (int i = 1; i <= n;i++)
T[i] = update(T[i-1], 1, m, a[i]);
long long ans=0,t=0;
for (int i=1;i<=n;++i) ans+=getsum1(i+1,n,a[i]);
printf("%lld",ans);ansS.insert(ans);
S.insert((node){1,n,ans});
for (int i=1;i<=n;++i){
scanf("%lld",&t); t=ans^t;
set<node>::iterator it=S.upper_bound((node){t,n+1,0});
it--;S.erase(it);
int l=(*it).l,r=(*it).r;long long pre_ans=(*it).ans;
ansS.erase(ansS.find(pre_ans));
long long ans_1=0,ans_2=0,sum=0;
if (t-l<r-t){
for (int i=l;i<=t-1;++i) ans_1+=getsum1(i+1,t-1,a[i]),sum+=getsum1(t+1,r,a[i]);
sum+=getsum2(l,t-1,a[t])+getsum1(t+1,r,a[t]);
ans_2=pre_ans-ans_1-sum;
} else {
for (int i=t+1;i<=r;++i) ans_2+=getsum1(i+1,r,a[i]),sum+=getsum2(l,t-1,a[i]);
sum+=getsum2(l,t-1,a[t])+getsum1(t+1,r,a[t]);
ans_1=pre_ans-ans_2-sum;
}
ansS.insert(ans_1),ansS.insert(ans_2);ans=*ansS.rbegin();
if (t>l) S.insert((node){l,t-1,ans_1});
if (t<r) S.insert((node){t+1,r,ans_2});
if (i!=n) printf(" %lld",ans);
}
puts("");
}
int main(){
int T;scanf("%d",&T);
while (T--) solve();
return 0;
}
这程序好像有点Bug,我给组数据试试?
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 10072kb
input:
3 5 4 3 1 1 1 5 4 5 3 1 10 9 7 1 4 7 8 5 7 4 8 21 8 15 5 9 2 4 5 10 6 15 4 8 8 1 12 1 10 14 7 14 2 9 13 10 3 37 19 23 15 7 2 10 15 2 13 4 5 8 7 10
output:
7 0 0 0 0 20 11 7 2 0 0 0 0 0 0 42 31 21 14 14 4 1 1 1 0 0 0 0 0 0
result:
ok 3 lines
Test #2:
score: 0
Accepted
time: 1478ms
memory: 34448kb
input:
11116 10 10 5 10 3 6 4 8 5 9 8 31 27 24 11 12 3 0 2 3 1 10 8 2 7 2 8 10 1 10 9 10 6 5 2 13 2 1 0 1 3 1 10 7 10 7 6 1 3 10 6 7 9 21 18 10 1 6 5 4 8 9 10 10 2 10 4 8 8 5 7 2 6 7 20 10 9 1 15 0 4 2 9 7 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 10 1 2 3 4 5 6 7 8 9 10 6 3 5 2 7 10 9 1 4 8 10 1 10 1 3...
output:
21 18 16 12 10 6 4 1 1 0 12 12 10 10 4 4 4 2 1 0 20 16 9 5 3 3 3 0 0 0 22 14 8 8 5 5 2 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 19 12 7 4 4 2 2 1 0 0 20 18 8 3 1 1 0 0 0 0 45 21 21 10 3 3 3 0 0 0 17 11 8 2 1 1 1 0 0 0 13 4 1 0 0 0 0 0 0 0 29 27 22 15 9 7 4 3 1 0 26 16 9 2 1 1 1 1 1 0 0 0 0 0 0 ...
result:
ok 11116 lines
Extra Test:
score: 0
Extra Test Passed