![Codeforces Round #440 (Div. 2, based on Technocup 2018 Elimination Round 2) C. Maximum splitting Codeforces Round #440 (Div. 2, based on Technocup 2018 Elimination Round 2) C. Maximum splitting](https://image.shishitao.com:8440/aHR0cHM6Ly9ia3FzaW1nLmlrYWZhbi5jb20vdXBsb2FkL2NoYXRncHQtcy5wbmc%2FIQ%3D%3D.png?!?w=700&webp=1)
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题目:
2 seconds
256 megabytes
standard input
standard output
You are given several queries. In the i-th query you are given a single positive integer ni. You are to represent ni as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.
The first line contains single integer q (1 ≤ q ≤ 105) — the number of queries.
q lines follow. The (i + 1)-th line contains single integer ni (1 ≤ ni ≤ 109) — the i-th query.
For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.
1
12
3
2
6
8
1
2
3
1
2
3
-1
-1
-1
12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.
8 = 4 + 4, 6 can't be split into several composite summands.
1, 2, 3 are less than any composite number, so they do not have valid splittings.
思路:
最小合数是4,所以用4去凑就行了
#include <bits/stdc++.h> using namespace std; #define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-;
const double pi=acos(-1.0);
const int K=1e6+;
const int mod=1e9+; int main(void)
{
int n,x,ans;cin>>x;
while(x--)
{
scanf("%d",&n);
if(n%==)
ans=n/;
else if(n%==)
{
if(n<)
ans=-;
else if(n==)
ans=;
else
ans=(n-)/+;
}
else if(n%==)
{
if(n<)
ans=-;
else if(n==)
ans=;
else
ans=n/;
}
else
{
if(n<)
ans=-;
else
ans=(n-)/+;
}
printf("%d\n",ans);
}
return ;
}