QOJ.ac
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 |
---|---|---|---|---|---|---|---|---|---|
#41698 | #3837. The Matching System | MCPlayer542 | WA | 1ms | 1848kb | C++ | 722b | 2022-07-31 15:23:07 | 2022-07-31 15:23:08 |
Judging History
answer
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,f[1010],md=1e9+7;
char s[1010],p[1010];
int main()
{
scanf("%d",&n);
if(n==1)
{
printf("0\n0\n1\n*\n0\n2\n");
return 0;
}
if(n==2)
{
printf("*0\n00\n3\n*0\n00\n4");
return 0;
}
if(n==4)
{
printf("**1*\n1000\n31\n**00\n0000\n10\n");
return 0;
}
for(int i=0;i<n;++i) s[i]='0',p[i]='*';
s[0]=p[n-2]='1';
for(int i=0;i<=n;++i) f[i]=1;
for(int i=0;i<n-2;++i)
for(int j=1;j<=n;++j)
f[j]=(f[j-1]+f[j])%md;
printf("%s\n%s\n%d\n",p,s,f[n]*2);
for(int i=0;i<n;++i) s[i]=p[i]='0';
for(int i=0;i<n-n/2;++i) p[i]='*';
printf("%s\n%s\n%d\n",p,s,(n/2+3)*(n/2)+(n&1)*(n/2+2));
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 1784kb
input:
3
output:
*1* 100 8 **0 000 7
result:
ok Accepted
Test #2:
score: 0
Accepted
time: 1ms
memory: 1804kb
input:
1
output:
0 0 1 * 0 2
result:
ok Accepted
Test #3:
score: 0
Accepted
time: 1ms
memory: 1768kb
input:
2
output:
*0 00 3 *0 00 4
result:
ok Accepted
Test #4:
score: 0
Accepted
time: 1ms
memory: 1848kb
input:
4
output:
**1* 1000 31 **00 0000 10
result:
ok Accepted
Test #5:
score: -100
Wrong Answer
time: 1ms
memory: 1828kb
input:
5
output:
***1* 10000 112 ***00 00000 14
result:
wrong answer The value of enery cost in the maximum mathing method is wrong