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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#492927 | #4912. WereYouLast | lifan | 100 ✓ | 5512ms | 4992kb | C++14 | 936b | 2024-07-26 17:15:38 | 2024-07-26 17:15:42 |
Judging History
answer
//
#include<bits/stdc++.h>
using namespace std;
bool query(int);
void modify(int,bool);
bool WereYouLast(int n,int m){
/*
其实就是用5个bit位存储当前的位置(2^k)
然后再对于每个位置维护这个位置出现过奇数/偶数次
如果是偶数次停留,否则回到0继续走
感觉这个数学分析之后好神奇
*/
//特判n=10;
if(n==1024||m==10){
int now=0;
for(int i=1;i<=10;i++)now<<=1,now|=query(i);
if(now==n-1)return 1;
++now;
for(int i=10;i;--i)modify(i,now&1),now>>=1;
return 0;
}
int pw=-1,tmp=n;
while(tmp)pw++,tmp>>=1;
int now=0;
for(int i=1;i<=5;i++)now<<=1,now|=query(i);
if(now==pw)return 1;
if(now==pw-1){//临门一脚
for(int i=5;i;--i)modify(i,pw&1),pw>>=1;
return 0;
}
int op=query(now+50);
if(op){
modify(50+now,0);
++now;
}
else modify(50+now,1),now=0;
for(int i=5;i;--i)modify(i,now&1),now>>=1;
return 0;
}
Details
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Subtask #1:
score: 10
Accepted
Test #1:
score: 10
Accepted
time: 1ms
memory: 3904kb
input:
1024 10
output:
12345876 10 10
result:
ok Correct Answer. C1 = 10. C2 = 10.
Subtask #2:
score: 20
Accepted
Test #2:
score: 20
Accepted
time: 6ms
memory: 4992kb
input:
65536 100000
output:
12345876 6 6
result:
ok Correct Answer. C1 = 6. C2 = 6.
Subtask #3:
score: 30
Accepted
Test #3:
score: 30
Accepted
time: 83ms
memory: 4976kb
input:
1048576 100000
output:
12345876 6 6
result:
ok Correct Answer. C1 = 6. C2 = 6.
Subtask #4:
score: 40
Accepted
Test #4:
score: 40
Accepted
time: 5512ms
memory: 4988kb
input:
67108864 100000
output:
12345876 6 6
result:
ok Correct Answer. C1 = 6. C2 = 6.