P.S.:
输出换行
三个方法
1.直接按照要求做
根据给的数，需要push,pop哪些数据，具有唯一性
数最多进栈一次，出栈一次
O(n)
1 Source Code 2 Problem: 1363 User: congmingyige 3 Memory: 728K Time: 63MS 4 Language: G++ Result: Accepted 5 6 Source Code 7 8 #include <cstdio> 9 #include <cstdlib> 10 #include <cmath> 11 #include <cstring> 12 #include <string> 13 #include <algorithm> 14 #include <iostream> 15 using namespace std; 16 #define ll long long 17 18 const double eps=1e-8; 19 const ll inf=1e9; 20 const ll mod=1e9+7; 21 const int maxn=1e5+10; 22 23 int a[maxn]; 24 int st[maxn]; 25 26 int main() 27 { 28 int n,i,j,g; 29 bool line=0; 30 while (1) 31 { 32 scanf("%d",&n); 33 if (n==0) 34 break; 35 36 if (!line) 37 line=1; 38 else 39 printf(" "); 40 41 g=0; 42 while (1) 43 { 44 scanf("%d",&a[1]); 45 if (a[1]==0) 46 break; 47 for (i=2;i<=n;i++) 48 scanf("%d",&a[i]); 49 50 j=0; 51 for (i=1;i<=n;i++) 52 { 53 while (j<a[i]) 54 st[++g]=++j; 55 if (st[g]==a[i]) 56 g--; 57 else 58 break; 59 } 60 61 if (i==n+1) 62 printf("Yes "); 63 else 64 printf("No "); 65 } 66 } 67 return 0; 68 } 69 /* 70 6 71 1 3 2 4 6 5 72 1 4 3 5 2 6 73 1 3 6 5 2 4 74 5 4 3 2 1 6 75 5 3 2 1 6 4 76 1 3 5 4 6 2 77 1 3 6 2 5 4 78 79 4 80 1 2 3 4 81 1 2 4 3 82 1 3 2 4 83 1 3 4 2 84 1 4 2 3 85 1 4 3 2 86 2 1 3 4 87 2 1 4 3 88 2 3 1 4 89 2 3 4 1 90 2 4 1 3 91 2 4 3 1 92 3 1 2 4 93 3 1 4 2 94 3 2 1 4 95 3 2 4 1 96 3 4 1 2 97 3 4 2 1 98 4 1 2 3 99 4 1 3 2 100 4 2 1 3 101 4 2 3 1 102 4 3 1 2 103 4 3 2 1 104 105 2 3 1 4 106 */
2.
一个序列，数x在第y个位置，第y个位置之后的小于x的数，必须是越往右越小。
即当前的最大数假设为z，则小于z的数必须是z-1,z-2,..,1。
时间上为O(n)，但常数较大
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <iostream> 8 using namespace std; 9 #define ll long long 10 11 const double eps=1e-8; 12 const ll inf=1e9; 13 const ll mod=1e9+7; 14 const int maxn=1e5+10; 15 16 int a[maxn]; 17 bool vis[maxn]; 18 19 int main() 20 { 21 int n,i,j; 22 bool line=0; 23 while (1) 24 { 25 scanf("%d",&n); 26 if (n==0) 27 break; 28 29 if (!line) 30 line=1; 31 else 32 printf(" "); 33 34 while (1) 35 { 36 scanf("%d",&a[1]); 37 if (a[1]==0) 38 break; 39 for (i=2;i<=n;i++) 40 scanf("%d",&a[i]); 41 42 memset(vis,0,sizeof(vis)); 43 j=0; 44 for (i=1;i<=n;i++) 45 { 46 if (j>a[i]) 47 break; 48 else if (j<=a[i]) 49 j=a[i]-1; 50 51 vis[a[i]]=1; 52 while (vis[j]) 53 j--; 54 } 55 56 if (i==n+1) 57 printf("Yes "); 58 else 59 printf("No "); 60 } 61 } 62 return 0; 63 } 64 /* 65 6 66 1 3 2 4 6 5 67 1 4 3 5 2 6 68 1 3 6 5 2 4 69 5 4 3 2 1 6 70 5 3 2 1 6 4 71 1 3 5 4 6 2 72 1 3 6 2 5 4 73 74 8 75 1 3 6 8 7 5 4 2 76 77 4 78 1 2 3 4 79 1 2 4 3 80 1 3 2 4 81 1 3 4 2 82 1 4 2 3 83 1 4 3 2 84 2 1 3 4 85 2 1 4 3 86 2 3 1 4 87 2 3 4 1 88 2 4 1 3 89 2 4 3 1 90 3 1 2 4 91 3 1 4 2 92 3 2 1 4 93 3 2 4 1 94 3 4 1 2 95 3 4 2 1 96 4 1 2 3 97 4 1 3 2 98 4 2 1 3 99 4 2 3 1 100 4 3 1 2 101 4 3 2 1 102 103 2 3 1 4 104 */
previous数组
一个数跳出，使用previous数组，这个数不再被使用
O(n)
1 3 2 6 5 4 9 8 7
则
previous[4] 0
previous[6] 6->5->4->0 0
previous[7] 7->6->0
极限数据
1 2 3 4 5 6 7 8 9
按照上述方法O(n^2)
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <string> 6 #include <algorithm> 7 #include <iostream> 8 using namespace std; 9 #define ll long long 10 11 const double eps=1e-8; 12 const ll inf=1e9; 13 const ll mod=1e9+7; 14 const int maxn=1e5+10; 15 16 int a[maxn],pre[maxn]; 17 bool vis[maxn]; 18 19 int main() 20 { 21 int n,i,j,k; 22 bool line=0; 23 while (1) 24 { 25 scanf("%d",&n); 26 if (n==0) 27 break; 28 29 if (!line) 30 line=1; 31 else 32 printf(" "); 33 34 while (1) 35 { 36 scanf("%d",&a[1]); 37 if (a[1]==0) 38 break; 39 for (i=2;i<=n;i++) 40 scanf("%d",&a[i]); 41 42 memset(vis,0,sizeof(vis)); 43 j=0; 44 for (i=1;i<=n;i++) 45 { 46 if (j>a[i]) 47 break; 48 else if (j<=a[i]) 49 j=a[i]-1,k=j; 50 51 vis[a[i]]=1; 52 while (vis[j]) 53 { 54 if (pre[j]) 55 j=pre[j]; 56 else 57 j--; 58 } 59 if (j!=k) 60 pre[k]=j; 61 } 62 63 if (i==n+1) 64 printf("Yes "); 65 else 66 printf("No "); 67 } 68 } 69 return 0; 70 } 71 /* 72 6 73 1 3 2 4 6 5 74 1 4 3 5 2 6 75 1 3 6 5 2 4 76 5 4 3 2 1 6 77 5 3 2 1 6 4 78 1 3 5 4 6 2 79 1 3 6 2 5 4 80 81 8 82 1 3 6 8 7 5 4 2 83 84 4 85 1 2 3 4 86 1 2 4 3 87 1 3 2 4 88 1 3 4 2 89 1 4 2 3 90 1 4 3 2 91 2 1 3 4 92 2 1 4 3 93 2 3 1 4 94 2 3 4 1 95 2 4 1 3 96 2 4 3 1 97 3 1 2 4 98 3 1 4 2 99 3 2 1 4 100 3 2 4 1 101 3 4 1 2 102 3 4 2 1 103 4 1 2 3 104 4 1 3 2 105 4 2 1 3 106 4 2 3 1 107 4 3 1 2 108 4 3 2 1 109 110 2 3 1 4 111 */