Two progressions CodeForces - 125D (暴力)
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大意: 给定序列, 求划分为两个非空等差序列.
暴搜, 加个记忆化剪枝.
#include <iostream> #include <sstream> #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 = 3e4+10; #else const int N = 111; #endif int n, d1, d2, c[N], *f = c; map<int,bool> v1[N], v2[N]; vector<int> a, b; int dfs() { if (f==c+n) return !b.empty(); if (a.size()<2||*f-a.back()==d1&&!v1[f-c+1].count(d1)) { a.pb(*f++); if (a.size()>=2) d1=a[1]-a[0], v1[f-c][d1]=1; if (dfs()) return 1; --f, a.pop_back(); } if (b.size()<2||*f-b.back()==d2&&!v2[f-c+1].count(d2)) { b.pb(*f++); if (b.size()>=2) d2=b[1]-b[0], v2[f-c][d2]=1; if (dfs()) return 1; --f, b.pop_back(); } return 0; } int main() { scanf("%d", &n); REP(i,0,n-1) scanf("%d", c+i); if (dfs()) { for (auto &&t:a) printf("%d ", t);hr; for (auto &&t:b) printf("%d ", t);hr; } else puts("No solution"); }
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