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  • 2017-2018 ACM-ICPC, NEERC, Northern Subregional Contest

    A. Auxiliary Project

    完全背包。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long LL;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int n;
    int f[1000005];
    int g[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
    int main()
    {
    	freopen("auxiliary.in", "r", stdin); 
    	freopen("auxiliary.out", "w", stdout);
    	while(~scanf("%d",&n))
    	{
    		MS(f, -1);
    		f[0] = 0;
    		for(int i = 0; i <= n; ++i)
    		{
    			for(int j = 0; j < 10; ++j)
    			{
    				if(i - g[j] >= 0 && f[i - g[j]] >= 0)
    				{
    					gmax(f[i], f[i - g[j]] + j);
    				}
    			}
    		}
    		printf("%d
    ", f[n]);
    	}
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

    B. Boolean Satisfiability

    设$t$为出现过的变量个数,若同时存在某个变量以及其反变量,则答案为$2^t$,否则答案为$2^t-1$。

    #include<cstdio>
    #include<cstring>
    typedef long long ll;
    const int N=100010;
    bool v[N],neg[N];
    char s[N];
    int n,i;
    int main(){
    	freopen("boolean.in", "r", stdin); 
    	freopen("boolean.out", "w", stdout);
    	scanf("%s",s+1);
    	n=strlen(s+1);
    	for(i=1;i<=n;){
    		if(s[i]=='|')i++;
    		else if(s[i]=='~'){
    			neg[s[i+1]]=1;
    			i+=2;
    		}else{
    			v[s[i]]=1;
    			i++;
    		}
    	}
    	ll ans=1,flag=1;
    	for(i=1;i<N;i++){
    		if(v[i]||neg[i])ans*=2;
    		if(v[i]&&neg[i])flag=0;
    	}
    	printf("%I64d",ans-flag);
    }
    

      

    C. Consonant Fencity

    $O(2^{19})$枚举所有辅音字母的大小写即可。

    #include<cstdio>
    #include<cstring>
    typedef long long ll;
    const int N=1000010;
    char s[N];
    int n,i,j,mx,now,ans,S;
    bool is[26],big[26];
    int g[26][26],w[26][26],q[26],m;
    int main(){
    	freopen("consonant.in", "r", stdin); 
    	freopen("consonant.out", "w", stdout);
    	scanf("%s",s);
    	n=strlen(s);
    	for(i=1;i<n;i++){
    		g[s[i-1]-'a'][s[i]-'a']++;
    	}
    	is['a'-'a']=1;
    	is['e'-'a']=1;
    	is['i'-'a']=1;
    	is['o'-'a']=1;
    	is['u'-'a']=1;
    	is['w'-'a']=1;
    	is['y'-'a']=1;
    	for(i=0;i<26;i++)if(!is[i])q[m++]=i;
    	for(i=0;i<26;i++)for(j=0;j<26;j++){
    		if(is[i]||is[j])g[i][j]=0;
    	}
    	for(i=0;i<m;i++)for(j=0;j<m;j++)w[i][j]=g[q[i]][q[j]];
    	for(S=0;S<1<<m;S++){
    		now=0;
    		for(i=0;i<m;i++)for(j=0;j<m;j++)if(((S>>i)^(S>>j))&1)now+=w[i][j];
    		if(now>mx)mx=now,ans=S;
    	}
    	//printf("mx=%d
    ",mx);
    	S=ans;
    	for(i=0;i<26;i++)big[i]=0;
    	for(i=0;i<m;i++)if(S>>i&1)big[q[i]]=1;
    	for(i=0;i<n;i++)if(big[s[i]-'a'])putchar(s[i]-'a'+'A');else putchar(s[i]);
    }
    

      

    D. Dividing Marbles

    留坑。

    E. Equal Numbers

    令$goal=lcm(a_1,a_2,...,a_n)$,那么对于每种数,可以变成另一个存在的倍数,或者直接变成$goal$。

    按照代价从小到大合并即可。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long LL;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 1e6 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int n;
    map<int, int>mop;
    vector<int>vt[2];
    int f[N];
    int main()
    {
    	freopen("equal.in", "r", stdin); 
    	freopen("equal.out", "w", stdout);
    	while(~scanf("%d",&n))
    	{
    		mop.clear();
    		for(int i = 1; i <= n; ++i)
    		{
    			int x;
    			scanf("%d", &x);
    			++mop[x];
    		}
    		int top = 1e6;
    		for(int i = 0; i <= 1; ++i)vt[i].clear();
    		for(auto it : mop)
    		{
    			int i = it.first;
    			vt[0].push_back(it.second);
    			for(int j = i + i; j <= top; j += i)if(mop.count(j))
    			{
    				vt[1].push_back(it.second);
    				break;
    			}
    		}
    		int ans = mop.size();
    		MS(f, 63); f[0] = ans;
    		
    		int sum = 0;
    		int sz = vt[1].size();
    		sort(vt[1].begin(), vt[1].end());
    		for(int j = 0; j < sz; ++j)
    		{
    			sum += vt[1][j];
    			gmin(f[sum], ans - j - 1);
    		}
    		
    		sum = 0;
    		sz = vt[0].size();
    		sort(vt[0].begin(), vt[0].end());
    		for(int j = 0; j < sz; ++j)
    		{
    			sum += vt[0][j];
    			gmin(f[sum], ans - j);
    		}
    		
    		for(int i = 0; i <= n; ++i)
    		{
    			if(i)gmin(f[i], f[i - 1]);
    			printf("%d ", f[i]);
    		}
    		puts("");
    	}
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    6
    3 4 1 2 1 2
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

    F. Fygon 2.0

    建立有向图,边$a ightarrow b$表示$aleq b$,那么每个SCC中的变量都要相等。

    缩完点之后得到一个$n$个点的DAG,那么在渐进意义下,去掉等号时间复杂度不变,总复杂度为$n!$,而实际复杂度为拓扑序的个数,状压DP即可。

    时间复杂度$O(n2^n)$。

    #include<cstdio>
    typedef long long ll;
    const int N=50;
    int n,m,i,j,k;
    int g[N][N],f[N],e[N];
    char s[100];
    int vis[500],mark[N];
    ll dp[1<<20];
    inline int id(char x){
    	if(x=='1')return -1;
    	if(x=='n')return -1;
    	if(vis[x]==-1)vis[x]=m++;
    	return vis[x];
    }
    inline void add(int x,int y){//x<=y
    	if(x<0||y<0)return;
    	g[x][y]=1;
    }
    int F(int x){return f[x]==x?x:f[x]=F(f[x]);}
    inline void merge(int x,int y){
    	if(F(x)!=F(y))f[f[x]]=f[y];
    }
    ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
    int main(){
    	freopen("fygon20.in", "r", stdin); 
    	freopen("fygon20.out", "w", stdout);
    	scanf("%d",&n);
    	n--;
    	for(i=0;i<500;i++)vis[i]=-1;
    	while(n--){
    		scanf("%s",s);//for
    		scanf("%s",s);
    		int A=id(s[0]);
    		scanf("%s",s);//in
    		scanf("%s",s);
    		int L=id(s[6]);
    		scanf("%s",s);
    		int R=id(s[0]);
    		add(L,A);
    		add(A,R);
    	}
    	for(k=0;k<m;k++)for(i=0;i<m;i++)for(j=0;j<m;j++)g[i][j]|=g[i][k]&&g[k][j];
    	for(i=0;i<m;i++)f[i]=i;
    	for(i=0;i<m;i++)for(j=0;j<m;j++)if(g[i][j]&&g[j][i])merge(i,j);
    	for(i=0;i<m;i++)mark[i]=-1;
    	n=0;
    	for(i=0;i<m;i++)if(mark[F(i)]<0){
    		mark[f[i]]=n++;
    	}
    	for(i=0;i<m;i++)for(j=0;j<m;j++)if(mark[f[i]]!=mark[f[j]]&&g[i][j])e[mark[f[i]]]|=1<<mark[f[j]];
    	dp[0]=1;
    	for(i=0;i<1<<n;i++)if(dp[i])for(j=0;j<n;j++)if(!(i>>j&1)&&!(e[j]&i))dp[i|(1<<j)]+=dp[i];
    	ll U=dp[(1<<n)-1],D=1;
    	for(i=2;i<=n;i++)D*=i;
    	ll gc=gcd(U,D);
    	U/=gc,D/=gc;
    	printf("%d %lld/%lld",n,U,D);
    }
    /*
    2
    for i in range(1, n):
    lag
      
    4
    for i in range(1, n):
    for j in range(1, i):
    for k in range(j, j):
    lag
    
    4
    for i in range(1, n):
    for j in range(1, i):
    for k in range(i, j):
    lag
    */
    

      

    G. Grand Test

    求出DFS树,对于一条非树边$(u,v)$,暴力将$u$到$v$路径上的树边染上这条非树边的颜色。

    若一条树边被染了两次色,则说明对应的两个简单环有公共边。

    仅保留两个简单环,任取两个度数至少为$3$的点作为起点和终点,然后爆搜出所有路径即可,一定恰好有$3$条简单路径。

    时间复杂度$O(n+m)$。

    #include<cstdio>
    #include<algorithm>
    #include<set>
    using namespace std;
    typedef pair<int,int>P;
    const int N=100010,M=200010;
    int Case,n,m,i,x,y,g[N],v[M<<1],nxt[M<<1],ed;
    int vis[N],dfn,flag,f[N],d[N];
    P col[N],A,B;
    int S,T,p[N];
    set<P>e;
    inline void add(int x,int y){
    	d[x]++;
    	v[++ed]=y;nxt[ed]=g[x];g[x]=ed;
    }
    void dfs(int x,int y){
    	f[x]=y;
    	vis[x]=++dfn;
    	for(int i=g[x];i;i=nxt[i]){
    		int u=v[i];
    		if(u==y)continue;
    		if(!vis[u]){
    			dfs(u,x);
    		}else if(vis[u]<vis[x]){
    			int j=x;
    			if(flag)continue;
    			while(j!=u){
    				if(col[j].first){
    					flag=1;
    					A=P(x,u);//down up
    					B=col[j];
    					break;
    				}
    				col[j]=P(x,u);
    				j=f[j];
    			}
    		}
    	}
    }
    inline void push(int x,int y){
    	if(x>y)swap(x,y);
    	e.insert(P(x,y));
    }
    inline void go(int x,int y){
    	push(x,y);
    	while(x!=y){
    		push(x,f[x]);
    		x=f[x];
    	}
    }
    void dfs2(int x,int y,int z){
    	p[z]=x;
    	if(x==T){
    		printf("%d",z);
    		for(int i=1;i<=z;i++)printf(" %d",p[i]);
    		puts("");
    		return;
    	}
    	for(int i=g[x];i;i=nxt[i])if(v[i]!=y)dfs2(v[i],x,z+1);
    }
    void solve(){
    	scanf("%d%d",&n,&m);
    	
    	for(i=1;i<=n;i++)g[i]=vis[i]=f[i]=0;
    	ed=dfn=flag=0;
    	for(i=1;i<=n;i++)col[i]=P(0,0);
    	
    	while(m--){
    		scanf("%d%d",&x,&y);
    		add(x,y);
    		add(y,x);
    	}
    	for(i=1;i<=n;i++)if(!vis[i]){
    		dfs(i,0);
    	}
    	if(!flag){
    		puts("-1");
    		return;
    	}
    	e.clear();
    	go(A.first,A.second);
    	go(B.first,B.second);
    	for(i=1;i<=n;i++)g[i]=d[i]=0;
    	ed=0;
    	for(set<P>::iterator it=e.begin();it!=e.end();it++){
    		x=it->first;
    		y=it->second;
    		add(x,y);
    		add(y,x);
    	}
    	S=T=0;
    	for(i=1;i<=n;i++)if(d[i]>2){
    		if(!S)S=i;
    		else T=i;
    	}
    	printf("%d %d
    ",S,T);
    	dfs2(S,0,1);
    }
    int main(){
    	freopen("grand.in", "r", stdin); 
    	freopen("grand.out", "w", stdout);
    	scanf("%d",&Case);
    	while(Case--)solve();
    }
    /*
    6 6
    3 6
    3 4
    1 4
    1 2
    1 3
    2 3
    */
    

      

    H. Hidden Supervisors

    贪心求出每个连通块的最大匹配、根的匹配情况以及内部还未匹配的点数。

    对于所有根已经匹配的连通块,将其直接连到$1$上最优。

    对于剩下的连通块,按内部未匹配点数从大到小依次贪心连边即可。

    时间复杂度$O(nlog n)$。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long LL;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int n;
    int fa[N];
    int match[N];
    vector<int>son[N];
    vector<int>nomatch;
    int ANS;
    void dfs(int x)
    {
    	match[x] = 0;
    	for(auto y : son[x])
    	{
    		dfs(y);
    		if(!match[y] && !match[x])
    		{
    			match[y] = x;
    			match[x] = y;
    			++ANS;
    		}
    		if(!match[y])nomatch.push_back(y);
    	}
    }
    struct A
    {
    	int sz;
    	int rt;
    	vector<int>vt;
    	bool operator < (const A & b)const
    	{
    		return sz > b.sz;
    	}
    }a[N];
    int main()
    {
    	freopen("hidden.in", "r", stdin); 
    	freopen("hidden.out", "w", stdout);
    	while(~scanf("%d",&n))
    	{
    		vector<int>rt;
    		rt.push_back(1);
    		for(int i = 1; i <= n; ++i)
    		{
    			son[i].clear();
    		}
    		for(int i = 2; i <= n; ++i)
    		{
    			scanf("%d", &fa[i]);
    			if(fa[i] == 0)
    			{
    				rt.push_back(i);
    			}
    			else
    			{
    				son[fa[i]].push_back(i);
    			}
    		}
    		
    		ANS = 0;
    		int rtsz = rt.size();
    		int g = 0;
    		vector<int>one;
    		for(int i = 0; i < rtsz; ++i)
    		{
    			nomatch.clear();
    			int x = rt[i];
    			dfs(x);
    			if(x == 1 || match[x])
    			{
    				fa[x] = 1;
    				for(auto y : nomatch)
    				{
    					one.push_back(y);
    				}
    				if(x == 1 && !match[1])
    				{
    					one.push_back(1);
    				}
    			}
    			else
    			{
    				++g;
    				a[g].sz = nomatch.size();
    				a[g].rt = x;
    				a[g].vt = nomatch;
    			}
    		}
    		sort(a + 1, a + g + 1);
    		//
    		//printf("treenum = %d
    ", g);
    		//
    		for(int i = 1; i <= g; ++i)
    		{
    			int x = a[i].rt;
    			if(one.size())
    			{
    				int ff = one.back();
    				fa[x] = ff;
    				one.pop_back();
    				++ANS;
    			}
    			else
    			{
    				fa[x] = 1;
    				one.push_back(x);
    			}
    			for(auto y : a[i].vt)
    			{
    				one.push_back(y);
    			}
    		}
    		printf("%d
    ", ANS);
    		for(int i = 2; i <= n; ++i)printf("%d ", fa[i]);
    		puts("");
    	}
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

    I. Intelligence in Perpendicularia

    答案$=$包围盒周长$-$图形周长。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long LL;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 1e3 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int n;
    const LL INF = 1e9;
    LL maxx, minx, maxy, miny;
    
    struct A
    {
    	LL x, y;
    }a[N];
    int main()
    {
    	freopen("intel.in", "r", stdin); 
    	freopen("intel.out", "w", stdout);
    	scanf("%d", &n);
    	maxx = -INF, maxy = -INF, minx = INF, miny = INF;
    	for(int i = 0; i < n; i ++){
    		scanf("%lld%lld", &a[i].x, &a[i].y);
    		gmax(maxx, a[i].x);
    		gmax(maxy, a[i].y);
    		gmin(minx, a[i].x);
    		gmin(miny, a[i].y);
    	}a[n] = a[0];
    	LL ans = 0;
    	for(int i = 0; i < n; i ++){
    		ans += abs(a[i].x - a[i + 1].x) + abs(a[i].y - a[i + 1].y);
    	}
    	ans -= (maxx - minx) * 2 + (maxy - miny) * 2;
    	printf("%lld
    ", ans);
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

    J. Joker

    分块维护凸壳。

    K. Kotlin Island

    枚举行列分别切了几刀即可。

    #include<stdio.h>
    #include<iostream>
    #include<string.h>
    #include<string>
    #include<ctype.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<algorithm>
    #include<time.h>
    using namespace std;
    void fre() {  }
    #define MS(x, y) memset(x, y, sizeof(x))
    #define ls o<<1
    #define rs o<<1|1
    typedef long long LL;
    typedef unsigned long long UL;
    typedef unsigned int UI;
    template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
    template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
    const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
    template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
    int casenum, casei;
    int n, m, g;
    char s[105][105];
    bool solve()
    {
    	MS(s, 0);
    	int y = (n - 1) / 2;
    	int x = (m - 1) / 2;
    	for(int i = 0; i <= y; ++i)
    	{
    		for(int j = 0; j <= x; ++j)
    		{
    			if( (i + 1) * (j + 1) == g) 
    			{
    				for(int ii = 1; ii <= n; ++ii)
    				{
    					for(int jj = 1; jj <= m; ++jj)
    					{
    						if(ii % 2 == 0 && ii <= i * 2 || jj % 2 == 0 && jj <= j * 2)
    						{
    							s[ii][jj] = '#';
    						}
    						else
    						{
    							s[ii][jj] = '.';
    						}
    					}
    				}
    				return 1;
    			}
    		}
    	}
    	return 0;
    }
    int main()
    {
    	freopen("kotlin.in", "r", stdin); 
    	freopen("kotlin.out", "w", stdout);
    	while(~scanf("%d%d%d",&n, &m, &g))
    	{
    		if(!solve())puts("Impossible");
    		else
    		{
    			for(int i = 1; i <= n; ++i)puts(s[i] + 1);
    		}
    	}
    	return 0;
    }
    
    /*
    【trick&&吐槽】
    
    
    【题意】
    
    
    【分析】
    
    
    【时间复杂度&&优化】
    
    
    */
    

      

    L. Little Difference

    若$n=2^k$形式则有无穷多个解,否则只能是$n=a^k$或$n=a^x(a+1)^y$的形式。

    枚举$x,y$后二分$a$即可。

    #include<cstdio>
    typedef long long ll;
    const ll lim=1000000000000000010LL;
    ll n;
    int ans;
    inline ll mul(ll a,ll b){
    	if(a>lim/b)return lim;
    	a*=b;
    	if(a>lim)a=lim;
    	return a;
    }
    inline ll po(ll a,int b){
    	ll t=1;
    	while(b--){
    		t=mul(t,a);
    		if(t>=lim)return lim;
    	}
    	return t;
    }
    inline ll get1(int k){
    	ll l=2,r=n,mid;
    	while(l<=r){
    		mid=(l+r)>>1;
    		ll t=po(mid,k);
    		if(t==n)return mid;
    		if(t<n)l=mid+1;else r=mid-1;
    	}
    	return 2;
    }
    inline ll get2(int i,int j){
    	ll l=2,r=n,mid;
    	while(l<=r){
    		mid=(l+r)>>1;
    		ll t=mul(po(mid,i),po(mid+1,j));
    		if(t==n)return mid;
    		if(t<n)l=mid+1;else r=mid-1;
    	}
    	return 2;
    }
    int main(){
    	freopen("little.in", "r", stdin); freopen("little.out", "w", stdout);
    	scanf("%lld",&n);
    	if(n<=2)return puts("-1"),0;
    	if(n==(n&-n))return puts("-1"),0;
    	//a^k
    	for(int _=0;_<2;_++){
    		for(int k=1;k<=70;k++){
    			ll t=get1(k);//t>=2
    			if(po(t,k)==n){
    				if(_==0)ans++;
    				else{
    					printf("%d",k);
    					for(int o=1;o<=k;o++)printf(" %lld",t);
    					puts("");
    				}
    			}
    		}
    		for(int i=1;i<=70;i++)for(int j=1;j<=70;j++){
    			ll t=get2(i,j);//t>=2
    			if(mul(po(t,i),po(t+1,j))==n){
    				if(_==0)ans++;
    				else{
    					printf("%d",i+j);
    					for(int o=1;o<=i;o++)printf(" %lld",t);
    					for(int o=1;o<=j;o++)printf(" %lld",t+1);
    					puts("");
    				}
    			}
    		}
    		if(_==0)printf("%d
    ",ans);
    	}
    }
    /*
    8589934592
    2176782336
    1000000000000000000
    */
    

      

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  • 原文地址:https://www.cnblogs.com/clrs97/p/7897895.html
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