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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#411833#6744. Squarehhu_yjh#AC ✓23ms3876kbC++17822b2024-05-15 20:23:192024-05-15 20:23:20

Judging History

你现在查看的是最新测评结果

  • [2024-05-15 20:23:20]
  • 评测
  • 测评结果:AC
  • 用时:23ms
  • 内存:3876kb
  • [2024-05-15 20:23:19]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int N=3e6+10;
const int mod=998244353;
ll X,Y;
int T;
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld",&X,&Y);
        if(X>=Y)
        {
            printf("%lld\n",X-Y);
            continue;
        }
        ll Nowx=sqrt(X*2);
        while((Nowx+1)*Nowx/2<X)
        ++Nowx;
        ll Nowy=sqrt(Y*2);
        while((Nowy+1)*Nowy/2<Y)
        ++Nowy;
        ll Posx=X-Nowx*(Nowx-1)/2;
        ll Posy=Y-Nowy*(Nowy-1)/2;
        // printf("%lld %lld\n%lld %lld\n",Nowx,Posx,Nowy,Posy);
        ll Ans=Nowy-Nowx;
        if(Nowx-Posx<=Nowy-Posy)
        Ans+=Nowy-Posy-Nowx+Posx;
        else Ans+=Posx+Nowy-Posy+1;
        printf("%lld\n",Ans);
    }
}

Details

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Test #1:

score: 100
Accepted
time: 1ms
memory: 3876kb

input:

2
5 1
1 5

output:

4
3

result:

ok 2 number(s): "4 3"

Test #2:

score: 0
Accepted
time: 23ms
memory: 3876kb

input:

100000
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
1 11
1 12
1 13
1 14
1 15
1 16
1 17
1 18
1 19
1 20
1 21
1 22
1 23
1 24
1 25
1 26
1 27
1 28
1 29
1 30
1 31
1 32
1 33
1 34
1 35
1 36
1 37
1 38
1 39
1 40
1 41
1 42
1 43
1 44
1 45
1 46
1 47
1 48
1 49
1 50
1 51
1 52
1 53
1 54
1 55
1 56
1 57
1 58
1 59
1 60
1 ...

output:

0
2
1
4
3
2
6
5
4
3
8
7
6
5
4
10
9
8
7
6
5
12
11
10
9
8
7
6
14
13
12
11
10
9
8
7
16
15
14
13
12
11
10
9
8
18
17
16
15
14
13
12
11
10
9
20
19
18
17
16
15
14
13
12
11
10
22
21
20
19
18
17
16
15
14
13
12
11
24
23
22
21
20
19
18
17
16
15
14
13
12
26
25
24
23
22
21
20
19
18
1
0
2
2
1
3
4
3
2
4
6
5
4
3
5
...

result:

ok 100000 numbers