标签：scanf ios inf std name ret line www. ide
https://www.luogu.org/problem/show?pid=T15744
无脑暴力，直接模拟就能水过
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<algorithm> 7 #define Ll long long 8 using namespace std; 9 const int MAXN=1011; 10 const int INF=0x7fffff; 11 inline int read() 12 { 13 char c=getchar();int flag=1,x=0; 14 while(c<‘0‘||c>‘9‘) {if(c==‘-‘) flag=-1;c=getchar();} 15 while(c>=‘0‘&&c<=‘9‘) x=x*10+c-48,c=getchar();return x*flag; 16 } 17 int n,p,k; 18 struct node 19 { 20 int l; 21 int change[MAXN]; 22 }lun[MAXN]; 23 int now[MAXN];//每一个位置的值 24 int pos[MAXN];//每一个值的位置 25 int main() 26 { 27 //freopen("rotate.in","r",stdin); 28 // freopen("rotate.out","w",stdout); 29 n=read();p=read();k=read(); 30 for(int i=1;i<=n;i++) now[i]=i,pos[i]=i;//每一个数现在是什么 31 for(int i=1;i<=p;i++) 32 { 33 lun[i].l=read(); 34 for(int j=1;j<=lun[i].l;j++) lun[i].change[j]=read(); 35 lun[i].l++;lun[i].change[lun[i].l]=lun[i].change[1]; 36 } 37 for(int i=p;i>=1;i--) 38 { 39 for(int j=1;j<=lun[i].l-1;j++) 40 now[ pos[ lun[i].change[j] ] ]= lun[i].change[j+1]; 41 for(int j=1;j<=n;j++) pos[now[j]]=j; 42 } 43 for(int i=1;i<=n;i++) 44 printf("%d ",now[i]); 45 46 return 0; 47 } 48 /* 49 (4,1)*(3,1)*(2,1)的话1变成2然后一直是2 50 2变成1然后变成3 51 3变成1然后变成4 52 4变成1 53 54 4 3 2 55 2 4 1 56 2 3 1 57 2 2 1 58 // 2 3 4 1 59 60 4 3 3 61 2 4 2 62 3 1 2 3 63 2 1 4 64 // 2 3 1 4 65 66 5 2 4 67 3 1 2 5 68 4 3 4 1 2 69 //5 3 4 2 1 70 */
不会，根据&和|的性质骗了60分
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<algorithm> 7 #define Ll long long 8 using namespace std; 9 const int MAXN=1e6+10; 10 const int INF=0x7fffff; 11 const int mod=1e9+7; 12 inline int read() 13 { 14 char c=getchar();int flag=1,x=0; 15 while(c<‘0‘||c>‘9‘) {if(c==‘-‘) flag=-1;c=getchar();} 16 while(c>=‘0‘&&c<=‘9‘) x=x*10+c-48,c=getchar();return x*flag; 17 } 18 int n,a,b,c,d; 19 int sz[MAXN]; 20 int ans=0; 21 int main() 22 { 23 // freopen("range.in","r",stdin); 24 // freopen("range.out","w",stdout); 25 n=read();a=read();b=read();c=read();d=read(); 26 for(int i=1;i<=n;i++) sz[i]=read(); 27 for(int i=1;i<=n;i++) 28 { 29 int yu=sz[i],huo=sz[i]; 30 for(int j=i;j<=n;j++) 31 { 32 yu=yu&sz[j]; 33 huo=huo|sz[j]; 34 if(yu>=a&&yu<=b&&huo>=c&&huo<=d) 35 ans=(ans+1)%mod; 36 if(yu<a||huo>d) break; 37 } 38 } 39 printf("%d",ans%mod); 40 return 0; 41 }
正解：
尝试固定左端点，观察右端点有什么性质
可以看出 and运算的值是递减的，or运算的值是递增的
用st表维护and和or的值||线段树||直接找结果
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <cmath> 5 #include <algorithm> 6 #include <cassert> 7 8 using namespace std; 9 typedef long long LL; 10 LL mod = 1e9+7; 11 LL cnt; 12 int s[100005],n,a,b,c,d; 13 int sta[100005][20],sto[100005][20]; 14 int aska(int l,int r){ 15 int L=r-l+1; 16 int t = log2(L); 17 return sta[l][t] & sta[r-(1<<t)+1][t]; 18 } 19 int asko(int l,int r){ 20 int L=r-l+1; 21 int t = log2(L); 22 return sto[l][t] | sto[r-(1<<t)+1][t]; 23 } 24 int main(){ 25 freopen("range.in","r",stdin); 26 freopen("range.out","w",stdout); 27 scanf("%d%d%d%d%d",&n,&a,&b,&c,&d); 28 for(int i=1;i<=n;i++) scanf("%d",&s[i]),sta[i][0]=sto[i][0]=s[i]; 29 // puts("aaa"); 30 for(int j=1;j<20;j++) 31 for(int i=1;i<=n;i++){ 32 if(i+(1<<j) -1 <=n){ 33 sta[i][j] = sta[i][j-1] & sta[i+(1<<(j-1))][j-1]; 34 sto[i][j] = sto[i][j-1] | sto[i+(1<<(j-1))][j-1]; 35 } 36 } 37 for(int i=1;i<=n;i++){ 38 int andsum = s[i], orsum = s[i]; 39 40 41 for(int j=i;j<=n;j++){ 42 int L=j,R=n+1; 43 andsum = aska(i,j); 44 orsum = asko(i,j); 45 while(R-L>1){ 46 int mid = (L+R)/2; 47 if(aska(i,mid) == andsum && asko(i,mid) == orsum){ 48 L=mid; 49 }else R=mid; 50 } 51 if(a <= andsum && andsum <= b && c <= orsum && orsum <= d){ 52 cnt += L-j+1; 53 } 54 j=L; 55 } 56 57 } 58 cout<<cnt%mod<<endl; 59 60 61 return 0; 62 }
不会，打30分暴力走人
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<algorithm> 7 #define Ll long long 8 using namespace std; 9 const int MAXN=1e6+10; 10 const int INF=0x7fffff; 11 const int mod=1e9+7; 12 inline int read() 13 { 14 char c=getchar();int flag=1,x=0; 15 while(c<‘0‘||c>‘9‘) {if(c==‘-‘) flag=-1;c=getchar();} 16 while(c>=‘0‘&&c<=‘9‘) x=x*10+c-48,c=getchar();return x*flag; 17 } 18 int n,k; 19 int a[MAXN]; 20 struct node 21 { 22 int u,v,w,f,nxt; 23 }edge[MAXN]; 24 int head[MAXN]; 25 int num=1; 26 inline void add_edge(int x,int y) 27 { 28 edge[num].u=x; 29 edge[num].v=y; 30 edge[num].nxt=head[x]; 31 head[x]=num++; 32 } 33 int ans=0; 34 int deep[MAXN]; 35 void pd() 36 { 37 queue<int>q;q.push(1); 38 int now=a[1]; 39 while(q.size()!=0) 40 { 41 int p=q.front();q.pop(); 42 for(int i=head[p];i!=-1;i=edge[i].nxt) 43 { 44 if(edge[i].f==0&&deep[edge[i].v]>deep[edge[i].u]) 45 { 46 now+=a[edge[i].v]; 47 q.push(edge[i].v); 48 } 49 } 50 } 51 if(now==k) ans=(ans+1)%mod; 52 } 53 void dfs(int now) 54 { 55 if(now==num) 56 { 57 pd(); 58 return ; 59 } 60 edge[now].f=1; 61 edge[now+1].f=1; 62 dfs(now+2); 63 edge[now].f=0; 64 edge[now+1].f=0; 65 dfs(now+2); 66 } 67 void make_deep(int now) 68 { 69 for(int i=head[now];i!=-1;i=edge[i].nxt) 70 if(deep[edge[i].v]==0) 71 deep[edge[i].v]=deep[now]+1,make_deep(edge[i].v); 72 } 73 int main() 74 { 75 //freopen("fruit.in","r",stdin); 76 // freopen("fruit.out","w",stdout); 77 memset(head,-1,sizeof(head)); 78 n=read();k=read(); 79 for(int i=1;i<=n;i++) a[i]=read(); 80 for(int i=1;i<=n-1;i++) 81 { 82 int x=read(),y=read(); 83 add_edge(x,y); 84 add_edge(y,x); 85 } 86 deep[1]=1; 87 make_deep(1); 88 edge[1].f=1;//断 89 edge[2].f=1; 90 dfs(3); 91 edge[1].f=0; 92 edge[2].f=0; 93 dfs(3);//没断 94 printf("%d",ans%mod); 95 return 0; 96 } 97 98 /* 99 5 3 100 2 1 0 1 1 101 1 2 102 1 3 103 3 4 104 3 5 105 //7 106 107 3 1 108 1 1 1 109 1 2 110 2 3 111 //2 112 */
f[i][j] 表示i 号节点在子树中拿j 个果子的方案数
DFS 时可以直接把父节点状态传下去，减少一维合并复杂度
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<algorithm> 7 #define Ll long long 8 using namespace std; 9 const int MAXN=1e6+10; 10 const int INF=0x7fffff; 11 const int mod=1e9+7; 12 inline int read() 13 { 14 char c=getchar();int flag=1,x=0; 15 while(c<‘0‘||c>‘9‘) {if(c==‘-‘) flag=-1;c=getchar();} 16 while(c>=‘0‘&&c<=‘9‘) x=x*10+c-48,c=getchar();return x*flag; 17 } 18 int n,k; 19 int a[MAXN]; 20 struct node 21 { 22 int u,v,w,f,nxt; 23 }edge[MAXN]; 24 int head[MAXN]; 25 int num=1; 26 inline void add_edge(int x,int y) 27 { 28 edge[num].u=x; 29 edge[num].v=y; 30 edge[num].nxt=head[x]; 31 head[x]=num++; 32 } 33 int ans=0; 34 int deep[MAXN]; 35 void pd() 36 { 37 queue<int>q;q.push(1); 38 int now=a[1]; 39 while(q.size()!=0) 40 { 41 int p=q.front();q.pop(); 42 for(int i=head[p];i!=-1;i=edge[i].nxt) 43 { 44 if(edge[i].f==0&&deep[edge[i].v]>deep[edge[i].u]) 45 { 46 now+=a[edge[i].v]; 47 q.push(edge[i].v); 48 } 49 } 50 } 51 if(now==k) ans=(ans+1)%mod; 52 } 53 void dfs(int now) 54 { 55 if(now==num) 56 { 57 pd(); 58 return ; 59 } 60 edge[now].f=1; 61 edge[now+1].f=1; 62 dfs(now+2); 63 edge[now].f=0; 64 edge[now+1].f=0; 65 dfs(now+2); 66 } 67 void make_deep(int now) 68 { 69 for(int i=head[now];i!=-1;i=edge[i].nxt) 70 if(deep[edge[i].v]==0) 71 deep[edge[i].v]=deep[now]+1,make_deep(edge[i].v); 72 } 73 int main() 74 { 75 //freopen("fruit.in","r",stdin); 76 // freopen("fruit.out","w",stdout); 77 memset(head,-1,sizeof(head)); 78 n=read();k=read(); 79 for(int i=1;i<=n;i++) a[i]=read(); 80 for(int i=1;i<=n-1;i++) 81 { 82 int x=read(),y=read(); 83 add_edge(x,y); 84 add_edge(y,x); 85 } 86 deep[1]=1; 87 make_deep(1); 88 edge[1].f=1;//断 89 edge[2].f=1; 90 dfs(3); 91 edge[1].f=0; 92 edge[2].f=0; 93 dfs(3);//没断 94 printf("%d",ans%mod); 95 return 0; 96 } 97 98 /* 99 5 3 100 2 1 0 1 1 101 1 2 102 1 3 103 3 4 104 3 5 105 //7 106 107 3 1 108 1 1 1 109 1 2 110 2 3 111 //2 112 */
总结
就喜欢这种题目区分度大的考试。
拿满暴力分就有一个不错的名次23333
标签：scanf ios inf std name ret line www. ide
原文地址：http://www.cnblogs.com/zwfymqz/p/7768823.html
