Codeforces Round #667 (Div. 3) ABCD
https://codeforces.com/contest/1409
A. Yet Another Two Integers Problem
题目大意：
k∈[1;10] 我们每次可以选择 a:=a+k or a:=a−k
问a要经历多少次操作变成b？
input
6
5 5
13 42
18 4
1337 420
123456789 1000000000
100500 9000
output
0
3
2
92
87654322
9150
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18;
const LL N=10000200,M=2002;
const LL mod=1e9+7;
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
LL a,b;
cin>>a>>b;
LL sum=abs(a-b);
LL res=sum/10;
if(sum%10!=0) res++;
cout<<res<<endl;
}
return 0;
}
B. Minimum Product
题目大意：
给定a,b,x,y和n，我们每次可以把a减掉一次，或者b减掉一次，最多总操作数是n次；
但是必须保证a>=x,b>=y；
问我们a*b的最大值是多少？
input
7
10 10 8 5 3
12 8 8 7 2
12343 43 4543 39 123212
1000000000 1000000000 1 1 1
1000000000 1000000000 1 1 1000000000
10 11 2 1 5
10 11 9 1 10
output
70
77
177177
999999999000000000
999999999
55
10
找规律即可
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18;
const LL N=10000200,M=2002;
const LL mod=1e9+7;
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
LL a,b,x,y,n;
cin>>a>>b>>x>>y>>n;
LL minn=MAXN;
LL a1=0,b1=0;
a1=max(x,a-n);
b1=max(y,b-(n-(a-a1)));
minn=min(minn,a1*b1);
//cout<<a1<<" "<<b1<<endl;
b1=max(y,b-n);
a1=max(x,a-(n-(b-b1)));
minn=min(minn,a1*b1);
//cout<<a1<<" "<<b1<<endl;
cout<<minn<<endl;
}
return 0;
}
C. Yet Another Array Restoration
题目大意：
给定数组总个数，以及数组中必定包含的两个数字
让我们构建出方差一致的数组a（保证数组a一定存在）。
input
5
2 1 49
5 20 50
6 20 50
5 3 8
9 13 22
output
1 49
20 40 30 50 10
26 32 20 38 44 50
8 23 18 13 3
1 10 13 4 19 22 25 16 7
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18;
const LL N=10000200,M=2002;
const LL mod=1e9+7;
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
LL n,x,y;
cin>>n>>x>>y;
if(n==2)
{
cout<<x<<" "<<y<<endl;
}
else
{
LL rest=y-x;
LL num=n-1;
while(rest%num!=0)
{
num--;
}
LL ad=rest/num;
//cout<<num<<" "<<ad<<endl;
LL sum=num+1;
vector<LL> v;
//cout<<"sum "<<sum<<endl;
if(n>sum)
{
LL idx=1;
while(x-idx*ad>=1&&n>sum)
{
v.push_back(x-idx*ad);
sum++;
idx++;
}
idx=1;
while(sum<n)
{
v.push_back(y+idx*ad);
sum++;
idx++;
}
}
for(LL i=x;i<=y;i+=ad)
{
v.push_back(i);
}
for(LL i=0;i<v.size();i++)
{
cout<<v[i]<<" ";
}
cout<<endl;
}
}
return 0;
}
D. Decrease the Sum of Digits
题目大意：
问我们这个数字n能够满足数位<=m的最小变化次数是多少？
每次变化就是+1。
input
5
2 1
1 1
500 4
217871987498122 10
100000000000000001 1
output
8
0
500
2128012501878
899999999999999999
abc有一个和这个好像的！！！
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18;
const LL N=5000200,M=2002;
const LL mod=1e9+7;
LL idx[N],sum[N];
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
cin>>T;
while(T--)
{
LL n,m;
cin>>n>>m;
for(int i=0;i<=18;i++)
{
if(i==0) idx[i]=1;
else idx[i]=idx[i-1]*10;
}
string s=to_string(n);
s=" "+s;
for(int i=1;i<s.size();i++)
{
sum[i]=sum[i-1]+(s[i]-'0');
}
LL odd=n;
LL now=n;
for(int i=s.size()-1,k=1;i>=1;i--,k++)
{
if(sum[i]>m)
{
now/=idx[k];
now*=idx[k];
now+=idx[k];
}
//cout<<now<<endl;
s=to_string(now);
s=" "+s;
for(int j=1;j<s.size();j++)
sum[j]=sum[j-1]+(s[j]-'0');
}
//cout<<now<<endl;
cout<<now-odd<<endl;
}
return 0;
}
