Codeforces Round #440 (Div. 2) C. Maximum splitting

C. Maximum splitting

Problem Statement

    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.

Input

    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.

Output

    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.

Examples

Example 1
    Input
        1
        12
    Output
        3
Example 2
    Input
        2
        6
        8
    Output
        1
        2
Example 3
    Input
        3
        1
        2
        3
    Output
        -1
        -1
        -1

Note

    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.

题意

    给出一个数，将这个数分成若干个合数的和，求最多能分成多少个合数

思路

    对于一个数，我们可以将它按照mod 4的意义下分类讨论：
    0 : 直接分成若干个4，ans = n / 4;
    1 : 提出一个9(模4余1意义下最小合数)，剩下的分成若干个4，ans = (n - 9) / 4 + 1
    2 : 提出一个6(模4余2意义下最小合数)，剩下的分成若干个4，ans = (n - 6) / 4 + 1
    3 : 提出一个15(模4余3意义下最小合数)，剩下的分成若干个4，ans = (n - 15) / 4 + 2

Code

#pragma GCC optimize(3)#include<bits/stdc++.h>using namespace std;typedef long long ll;inline void readInt(int &x) {    x=0;int f=1;char ch=getchar();    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();    x*=f;}inline void readLong(ll &x) {    x=0;int f=1;char ch=getchar();    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();    x*=f;}/*================Header Template==============*/int n,k;int main() {    cin>>n;    while(n--) {        cin>>k;        if(k==1||k==2||k==3||k==5||k==7||k==11) {            puts("-1");            continue;        }        if(k%4==0) {            printf("%d\n",k/4);            continue;        }        if(k%4==1) {            printf("%d\n",(k-9)/4+1);            continue;        }        if(k%4==2) {            printf("%d\n",k/4);            continue;        }        if(k%4==3) {            printf("%d\n",(k-9)/4+1);            continue;        }    }}