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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 |
---|---|---|---|---|---|---|---|---|---|
#377606 | #6185. Best Problem | Kevin5307 | TL | 0ms | 0kb | C++20 | 3.1kb | 2024-04-05 15:48:50 | 2024-04-05 15:48:52 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
bool memBeg;
template<typename T> void chkmin(T &x,T y) {if(x>y) x=y;}
template<typename T> void chkmax(T &x,T y) {if(x<y) x=y;}
constexpr int mod=1e9+7;
void inc(int &x,int y) {x+=y; x-=(x>=mod)*mod;}
void dec(int &x,int y) {x-=y; x+=(x<0)*mod;}
constexpr int add(int x,int y) {return (x+y>=mod)?x+y-mod:x+y;}
constexpr int sub(int x,int y) {return (x<y)?x-y+mod:x-y;}
constexpr int quick_pow(int x,int times) {
int ret=1;
for(;times;times>>=1,x=1ll*x*x%mod) {
if(times&1) ret=1ll*ret*x%mod;
}
return ret;
}
constexpr int maxn=5e6+15;
int n;
char s[maxn],t[maxn];
pair<int,char> stk[maxn];
ll ret,cot[maxn];
bool memEn;
void fl() {
freopen(".in","r",stdin);
freopen(".out","w",stdout);
}
void solve(){
// fprintf(stderr,"%.24lf\n",fabs(&memEn-&memBeg)/1024.0/1024.0);
// fl();
scanf("%s",s+1);
n=strlen(s+1);
int top=0;
ll ret=0ll;
// 0...1 -> .....
auto emplace=[&](int idx,char c) {
if(top&&stk[top].second=='0'&&c=='1'&&((idx-stk[top].first)&3)==0) {
ret+=(idx-stk[top].first)>>2; top--;
} else stk[++top]={idx,c};
};
emplace(0,'1');
if(s[1]=='1') emplace(1,'1');
for(int i=2;i<=n;i++) if(s[i]==s[i-1]) emplace(i,s[i]);
if(s[n]=='0') emplace(n+1,'0');
emplace(n+2,'0');
// 0.... -> ....0
for(int i=top-1;i>=1;i--) {
if(stk[i].second=='0'&&stk[i+1].second=='0') {
int t=(stk[i+1].first-stk[i].first)>>2;
ret+=t; stk[i].first+=t<<2;
}
}
// ....1 -> 1....
for(int i=2;i<=top;i++) {
if(stk[i].second=='1'&&stk[i-1].second=='1') {
int t=(stk[i].first-stk[i-1].first)>>2;
ret+=t; stk[i].first-=t<<2;
}
}
// printf("ret = %lld\n",ret);
for(int i=6;i<=n+2;i++) {
int t=(i-2)>>2;
cot[i]=t+cot[t<<2];
}
int ept[2]={-1,-1};
ll mx[2]={0ll,0ll};
for(int i=1,i0,i1;i<=top;i=i1) {
for(i0=i;i0<=top&&stk[i0].second=='0';i0++);
for(i1=i0;i1<=top&&stk[i1].second=='1';i1++);
if(i==1) {
ept[0]=ept[1]=stk[i1-1].first;
} else if(i1==top+1) {
for(int t=0;t<2;t++) mx[t]+=cot[stk[i].first-ept[t]];
} else {
int nept[2]={-1,-1};
ll nmx[2]={0ll,0ll};
int step=(stk[i0].first-stk[i0-1].first)>>2;
// 0.....1 -> ....0.1
nept[0]=stk[i1-1].first;
for(int t=0;t<2;t++) {
chkmax(nmx[0],mx[t]+cot[(stk[i].first+(step<<2))-ept[t]]+1ll*(i0-i)*step);
}
// 0.....1 -> 0.1....
nept[1]=stk[i1-1].first-(step<<2);
for(int t=0;t<2;t++) {
chkmax(nmx[1],mx[t]+cot[stk[i].first-ept[t]]+1ll*(i1-i0)*step);
}
memcpy(ept,nept,sizeof(int)*2);
memcpy(mx,nmx,sizeof(ll)*2);
}
}
printf("%lld\n",ret+max(mx[0],mx[1]));
}
int main()
{
int _;scanf("%d",&_);
while(_--)solve();
}
詳細信息
Test #1:
score: 0
Time Limit Exceeded
input:
10100010011001011111
output:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...