大意: 求將[1,n]劃分成兩個集合, 且兩集合的和的差盡量小.

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和/2為偶數(shù)最小差一定為0, 和/2為奇數(shù)一定為1.

顯然可以通過某個前綴和刪去一個數(shù)得到.

#include <iostream> #include <iostream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //head#ifdef ONLINE_JUDGE const int N = 1e6+10; #else const int N = 111; #endifint n, m, a[N]; char s[N];int main() {scanf("%d", &n);int s = (1+n)*n/2, t = 0;REP(i,1,n) {t += i;if (t>=s/2&&t-s/2<=i) {vector<int> g;int r = 0;REP(j,1,i) if (j!=t-s/2) g.pb(j),r+=j; int rr = s-r;printf("%d\n%d ", abs(r-rr),int(g.size()));for (auto &&t:g) printf("%d ",t);return hr,0;}} }

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轉(zhuǎn)載于:https://www.cnblogs.com/uid001/p/10763714.html

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