Integer Sequence Dividing CodeForces - 1102A (规律)
Posted qieqiemin
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You are given an integer sequence 1,2,…,n1,2,…,n. You have to divide it into two sets AAand BB in such a way that each element belongs to exactly one set and |sum(A)−sum(B)||sum(A)−sum(B)| is minimum possible.
The value |x||x| is the absolute value of xx and sum(S)sum(S) is the sum of elements of the set SS.
Input
The first line of the input contains one integer nn (1≤n≤2⋅1091≤n≤2⋅109).
Output
Print one integer — the minimum possible value of |sum(A)−sum(B)||sum(A)−sum(B)| if you divide the initial sequence 1,2,…,n1,2,…,n into two sets AA and BB.
Examples
3
0
5
1
6
1
Note
Some (not all) possible answers to examples:
In the first example you can divide the initial sequence into sets A={1,2}A={1,2} and B={3}B={3} so the answer is 00.
In the second example you can divide the initial sequence into sets A={1,3,4}A={1,3,4} and B={2,5}B={2,5} so the answer is 11.
In the third example you can divide the initial sequence into sets A={1,4,5}A={1,4,5} and B={2,3,6}B={2,3,6} so the answer is 11.
题意:给你一个整数N,让你将1~N这N个整数分成两个集合,
问这两个集合的元素数值和的差最小能是多少。
思路:
先写几个样例来看下。
当N=3,
1,2,3 可以把1和2分到一个集合,3分到另一个集合。这样差为0
当N=4
1,2,3,4可以把 1和4分到一个集合,2和3在另一个集合,这样差为0
当N=5
1,2,3,4,5,可以分成这样{1,3,4},{2,5} 差为1
我们在算下这三个样例的所有元素和
N=3 ,sum=6
N=4,sum=10
N=5,sum=15
规律就可以看出来了,当1~N的和为偶数的时候,一定可以分成两个相同的sum的集合
为奇数可以分成相差为1的两个集合。
那么就根据规律来写程序了。
我的AC代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), ‘ ‘, sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define gg(x) getInt(&x) using namespace std; typedef long long ll; inline void getInt(int* p); const int maxn=1000010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ ll n; int main() { cin>>n; ll ans=(n*(1+n))/2ll; if(ans&1) { cout<<1<<endl; }else { cout<<0<<endl; } return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ‘ ‘ || ch == ‘ ‘); if (ch == ‘-‘) { *p = -(getchar() - ‘0‘); while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 - ch + ‘0‘; } } else { *p = ch - ‘0‘; while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 + ch - ‘0‘; } } }
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