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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 |
---|---|---|---|---|---|---|---|---|---|
#411833 | #6744. Square | hhu_yjh# | AC ✓ | 23ms | 3876kb | C++17 | 822b | 2024-05-15 20:23:19 | 2024-05-15 20:23:20 |
Judging History
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);
}
}
詳細信息
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