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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
#595136#7757. Palm Islandforget-star#WA 1ms7652kbC++141.3kb2024-09-28 12:41:112024-09-28 12:41:23

Judging History

你现在查看的是最新测评结果

  • [2024-09-28 12:41:23]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:7652kb
  • [2024-09-28 12:41:11]
  • 提交

answer

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

#define ll long long

using namespace std;

const int N=1e6+7,inf = 2e9+77777777;

int read()
{
    int a = 0,x = 1;char ch = getchar();
    while(ch > '9' || ch < '0') {if(ch == '-') x = -1;ch = getchar();}
    while(ch >= '0' && ch <= '9') {a = a*10 + ch-'0';ch = getchar();}
    return a*x;
}

int n,b[N],p[N],o,ans[N];
struct node{
    int l,r,num;
}a[N];
int nxt() {return a[a[o].r].num;}
void op1() {o = a[o].r;}
void op2() 
{
    int xt1 = a[o].r;
    swap(a[o].num,a[xt1].num);
    op1();
}

void solve()
{
    n = read();o = 0;int cnt=  0;
    for(int i = 0;i < n;i ++) a[i].num = read(),a[i].l = (i+n-1)%n,a[i].r = (i+1)%n;
    for(int i = 0;i < n;i ++) b[i] = read();
    for(int i = 0;i < n;i ++) p[b[i]] = b[(i+1)%n];
    for(int i = 1;i < n;i ++) {
        if(nxt() != p[a[o].num]) {
            ans[++cnt] = 2;
            op2();
        } else {
            ans[++cnt] = 1;
            op1();
        }
    }
    while(a[o].num != b[0]) {
        ans[++cnt] = 1;
        op1();
    }
    for(int i= 1;i <= cnt;i ++) printf("%d",ans[i]);
    puts("");
}

int main()
{
    // freopen("in.in","r",stdin);freopen("out.out","w",stdout);
    int T = read();
    while(T --) solve();
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 7652kb

input:

2
3
1 2 3
2 3 1
4
1 2 3 4
2 1 3 4

output:

1111
2111

result:

ok Correct. (2 test cases)

Test #2:

score: -100
Wrong Answer
time: 0ms
memory: 7644kb

input:

200
3
3 1 2
2 3 1
4
2 4 1 3
2 1 4 3
4
1 4 2 3
2 1 3 4
5
4 3 2 1 5
2 4 5 3 1
5
2 1 5 4 3
5 2 4 1 3
4
4 3 1 2
1 2 4 3
3
1 2 3
3 1 2
4
1 4 2 3
2 1 4 3
4
1 3 2 4
1 4 3 2
3
3 2 1
1 3 2
3
2 3 1
1 3 2
4
1 4 3 2
3 1 2 4
3
1 2 3
1 3 2
3
3 2 1
2 3 1
5
5 1 3 2 4
2 4 5 1 3
4
4 3 1 2
1 4 3 2
4
1 3 4 2
2 4 3 1
3
...

output:

11
21211
22111
222111
221211
111111
11
12111
221111
11
21
22111
2111
211
11111111
12111
221
1111
1111
2112
121
22111
211
121111
2221111
1111
212111
122111
121
211111
2111
221
211
221111
2211
222111
211
22111
2211
111
221
11
2111111
12111
221111
22211111
2111
11
22111111
111
1211
1121111
211
22111
22...

result:

wrong answer On Case#2: After your operations, a[2] = 3 but a[2] = 1. (test case 2)