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 |
---|---|---|---|---|---|---|---|---|---|
#397033 | #6744. Square | xuezhu | AC ✓ | 40ms | 3696kb | C++14 | 863b | 2024-04-23 15:48:38 | 2024-04-23 15:48:39 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
#define pii pair<int,int>
#define mp make_pair
#define ll long long
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const ll nn=3e5+5;
ll t,x,y;
pii find(ll x){
ll p=(ll)(sqrt(2.0*x)+1.5)-1;//对应的+后面的数-1
ll rst=x-p*(p-1)/2;//第rst个
return mp(p,rst);
}
ll f1(pii x,pii y){//情况1
ll z=y.first-x.first;
x={x.first+z,x.second+z};//x做z次加法运算
if(x.second>=y.second) return z+x.second-y.second;
return 1e18;
}
ll f2(pii x,pii y){//情况2s
ll z=x.second;
x={x.first-1,x.first-1};//先减到前一个p的最后一个数字,再运算
return z+f1(x,y);
}
int main(){
IOS;
cin>>t;
while(t--){
cin>>x>>y;
if(x>=y) cout<<x-y<<endl;
else{
pii x1=find(x),x2=find(y);
cout<<min(f1(x1,x2),f2(x1,x2))<<endl;
}
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 3696kb
input:
2 5 1 1 5
output:
4 3
result:
ok 2 number(s): "4 3"
Test #2:
score: 0
Accepted
time: 40ms
memory: 3628kb
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