1001 Olympiad
写的时候直接统计的个数，后来才知道应该递推一下
1 #include <cstdio> 2 #include <ctime> 3 #include <cstring> 4 #include <cstdlib> 5 #include <cmath> 6 #include <vector> 7 #include <map> 8 #include <set> 9 #include <stack> 10 #include <queue> 11 #include <string> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 16 #define getmid(l,r) ((l) + ((r) - (l)) / 2) 17 #define MP(a,b) make_pair(a,b) 18 #define PB push_back 19 20 typedef long long ll; 21 typedef pair<int,int> pii; 22 const double eps = 1e-8; 23 const int INF = (1 << 30) - 1; 24 const int maxn = 100005; 25 int f[maxn]; 26 27 int vis[maxn]; 28 29 int ok(int x){ 30 int a[10]; 31 int cnt = 0; 32 while(x){ 33 a[cnt++] = x%10; 34 x = x/10; 35 } 36 int b[10]; 37 memset(b,0,sizeof(b)); 38 for(int i = 0;i < cnt;i++){ 39 if(!b[a[i]]) b[a[i]] = 1; 40 else return 0; 41 } 42 43 return 1; 44 } 45 46 int main(){ 47 memset(vis,0,sizeof(vis)); 48 for(int i = 1;i <= maxn;i++) if(ok(i)) vis[i] = 1; 49 // for(int i = 1;i <= 100;i++) printf("vis[%d] = %d ",i,vis[i]); 50 f[0] = 1; 51 for(int i = 1;i <= maxn;i++) { 52 if(ok(i)) f[i] = f[i-1] +1; 53 else f[i] = f[i-1]; 54 } 55 56 57 int T; 58 scanf("%d",&T); 59 while(T--){ 60 int a,b; 61 scanf("%d%d",&a,&b); 62 printf("%d ",f[b] - f[a-1]); 63 } 64 return 0; 65 }
1002 Problem Killer
可以尺取法，或者递推
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include <cmath> 5 #include<stack> 6 #include<vector> 7 #include<map> 8 #include<set> 9 #include<queue> 10 #include<algorithm> 11 using namespace std; 12 13 typedef long long LL; 14 const int INF = (1<<30)-1; 15 const int mod=1000000007; 16 const int maxn=1000005; 17 18 LL a[maxn]; 19 int n; 20 21 int main(){ 22 int T; 23 scanf("%d",&T); 24 while(T--){ 25 scanf("%d",&n); 26 for(int i = 0;i < n;i++) scanf("%I64d",&a[i]); 27 28 if(n <= 2){ 29 printf("%d ",n); 30 continue; 31 } 32 33 int res = 0,ans = 0; 34 for(int i = 0;i < n;){ 35 int j = i+1; 36 while(j+1 < n && a[j-1] + a[j+1] == 2*a[j]) j++; 37 res = max(res,j-i+1); 38 i = j; 39 // printf("j = %d res = %d ",j,res); 40 } 41 42 for(int i = 0;i < n;){ 43 int j = i+1; 44 while(j+1 < n && a[j-1] * a[j+1] == a[j] * a[j]) j++; 45 ans = max(ans,j-i+1); 46 i = j; 47 // printf("j = %d ans = %d ",j,ans); 48 } 49 printf("%d ",max(res,ans)); 50 } 51 return 0; 52 }
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 const int maxn = 1000005; 8 int dp1[maxn],dp2[maxn]; 9 int n; 10 long long a[maxn]; 11 12 int ok1(int x){ 13 if(2*a[x] == a[x-1] + a[x+1]) return 1; 14 return 0; 15 } 16 17 int ok2(int x){ 18 if(a[x] * a[x] == a[x-1] * a[x+1]) return 1; 19 return 0; 20 } 21 22 int main(){ 23 int T; 24 scanf("%d",&T); 25 while(T--){ 26 scanf("%d",&n); 27 for(int i = 1;i <= n;i++) scanf("%I64d",&a[i]); 28 if(n <= 2){ 29 printf("%d ",n); 30 continue; 31 } 32 for(int i = 1;i <= n;i++) dp1[i] = dp2[i] = 2; 33 34 for(int i = 2;i < n;i++) { 35 if(ok1(i)) dp1[i] = dp1[i-1] + 1; 36 if(ok2(i)) dp2[i] = dp2[i-1] + 1; 37 } 38 int ans = 2; 39 for(int i = 1;i <= n;i++) ans = max(ans,max(dp1[i],dp2[i])); 40 printf("%d ",ans); 41 } 42 return 0; 43 }
1004 Route Statistics
1005 Simple Problem
1006 Test for Rikka
1007 Undirected Graph
1009 Walk Out
先从起点开始尽量走0,然后找到一堆距离终点最近的1，看作一层的1，然后从这些1出发，一定是向右边或者是下面走的
因为第一个1的高位已经确定了，位数短的串比位数长的串小
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<queue> 6 #include<vector> 7 using namespace std; 8 9 char g[1005][1005]; 10 int n,m;//n行,m列 11 int vis[1005][1005]; 12 int dir[4][2] = {1,0,-1,0,0,1,0,-1}; 13 14 int ok(int x,int y){ 15 if(vis[x][y] || x < 1 || x > n || y < 1 || y > m) return 0; 16 return 1; 17 } 18 19 void bfs(){ 20 queue< pair<int, int> > q; 21 memset(vis,0,sizeof(vis));vis[1][1] = 1; 22 int mx = 2; 23 q.push(make_pair(1,1)); 24 while(!q.empty()){ 25 int x = q.front().first; 26 int y = q.front().second;q.pop(); 27 28 if(g[x][y] == '0'){ 29 for(int i = 0;i < 4;i++){ 30 int xx = x+dir[i][0]; 31 int yy = y+dir[i][1]; 32 if( !ok(xx,yy)) continue; 33 q.push(make_pair(xx,yy)) ; 34 mx = max(mx,xx+yy);vis[xx][yy] = 1; 35 } 36 } 37 } 38 39 if(vis[n][m] && g[n][m] == '0') { 40 puts("0"); 41 return ; 42 } 43 44 printf("1"); 45 46 for(int i = mx;i < n+m ;i++){ 47 char mn = '1'; 48 for(int x = 1;x <= n;x++){ 49 int y = i - x; 50 if(y < 1 ||y > m || !vis[x][y]) continue; 51 mn = min(mn,g[x+1][y]); 52 mn = min(mn,g[x][y+1]); 53 } 54 printf("%c",mn); 55 56 for(int x = 1;x <= n;x++){ 57 int y = i-x; 58 if(y < 1 || y > m || !vis[x][y]) continue; 59 if(g[x+1][y] == mn) vis[x+1][y] = 1; 60 if(g[x][y+1] == mn) vis[x][y+1] = 1; 61 } 62 } 63 puts(""); 64 } 65 66 int main(){ 67 int T; 68 scanf("%d",&T); 69 while(T--){ 70 scanf("%d %d",&n,&m); 71 for(int i = 1;i <= n;i++) scanf("%s",g[i] + 1); 72 for(int i = 1;i <= n;i++) g[i][m+1] = '2'; 73 for(int i = 1;i <= m;i++) g[n+1][i] = '2'; 74 bfs(); 75 } 76 return 0; 77 }
1010 XYZ and Drops
1012 ZZX and Permutations