题意: 对于一个数操作n次,操作如下: 如果末尾是0就将这个数除以10, 如果末尾不是0就将这个数-1, 直接做就好了。
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define max3(a,b,c) max(a,max(b,c)) 12 #define min3(a,b,c) min(a,min(b,c)) 13 typedef pair<int,int> pll; 14 const int INF = 0x3f3f3f3f; 15 const LL mod = 1e9+7; 16 const int N = 1e5+10; 17 int main(){ 18 ///Fopen; 19 int n, k; 20 scanf("%d%d", &n, &k); 21 while(k--){ 22 int t = n%10; 23 if(t) n--; 24 else n/=10; 25 } 26 printf("%d", n); 27 return 0; 28 }
题意:求2个连续字符出现次数最多的那2个字符。
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define max3(a,b,c) max(a,max(b,c)) 12 #define min3(a,b,c) min(a,min(b,c)) 13 typedef pair<int,int> pll; 14 const int INF = 0x3f3f3f3f; 15 const LL mod = 1e9+7; 16 const int N = 1e3+10; 17 int cnt[N][N]; 18 char str[N]; 19 int main(){ 20 ///Fopen; 21 int n; 22 scanf("%d", &n); 23 scanf("%s", str); 24 int ans = -1, s1, s2; 25 for(int i = 0; i < n-1; i++){ 26 int x = str[i]-'A'; 27 int y = str[i+1]-'A'; 28 cnt[x][y]++; 29 if(cnt[x][y] > ans){ 30 ans = cnt[x][y]; 31 s1 = x; 32 s2 = y; 33 } 34 } 35 printf("%c%c",s1+'A',s2+'A'); 36 return 0; 37 }
题意:给你一个数组, 要求找到一个数在[1,1e9]之间,使得恰好有k个数小于等于他,如果没有输出-1, 有就输出任意一个合法答案。
题解:sort一下数组, 如果第k个数等于第k+1个数就找不到一个数满足题意了。 还有0个的情况。
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define max3(a,b,c) max(a,max(b,c)) 12 #define min3(a,b,c) min(a,min(b,c)) 13 typedef pair<int,int> pll; 14 const int INF = 0x3f3f3f3f; 15 const LL mod = 1e9+7; 16 const int N = 2e5+10; 17 int n, m; 18 int A[N]; 19 int main(){ 20 ///Fopen; 21 scanf("%d%d", &n, &m); 22 for(int i = 1; i <= n; i++) 23 scanf("%d", &A[i]); 24 sort(A+1, A+1+n); 25 if(m == 0){ 26 if(A[1] == 1) printf("-1"); 27 else printf("1"); 28 } 29 else if(m == n){ 30 printf("%d", (int)1e9); 31 } 32 else{ 33 if(A[m] == A[m+1]) printf("-1"); 34 else printf("%d", A[m]); 35 } 36 return 0; 37 }
D Divide by three, multiply by two
题意:将题目给的数重新排列,使得前一个数是后一个数的3倍,或者后一个数是前一个数2倍的情况最多。
题解:这个特殊的数列是不可能形成一个环的。DFS找一下就可以了,从长的开始输出。(也有更简单的方法, 但是当时写了DFS就DFS了。。。)
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define max3(a,b,c) max(a,max(b,c)) 12 #define min3(a,b,c) min(a,min(b,c)) 13 typedef pair<int,int> pll; 14 const int INF = 0x3f3f3f3f; 15 const LL mod = 1e9+7; 16 const int N = 1e2+10; 17 int n, m; 18 LL A[N]; 19 int cnt[N]; 20 LL ans[N]; 21 map<LL,int> mp; 22 void dfs(LL u){ 23 cnt[mp[u]] = 1; 24 if(u%3 == 0 && mp.count(u/3)){ 25 if(cnt[mp[u/3]] == -1) 26 dfs(u/3); 27 cnt[mp[u]] = max(cnt[mp[u]],1+cnt[mp[u/3]]); 28 } 29 if(mp.count(u*2)){ 30 if(cnt[mp[u*2]] == -1) 31 dfs(u*2); 32 cnt[mp[u]] = max(cnt[mp[u]],1+cnt[mp[u*2]]); 33 } 34 } 35 void Show(LL u, int h){ 36 printf("%I64d ", u); 37 A[mp[u]] = 0; 38 h--; 39 if(h == 0) return ; 40 if(u%3 == 0 && mp.count(u/3) && cnt[mp[u/3]] == h && A[mp[u/3]] != 0) 41 Show(u/3,h); 42 43 else if(mp.count(u*2) 44 && A[mp[u*2]] != 0 45 && cnt[mp[u*2]] == h){ 46 Show(u*2,h); 47 } 48 } 49 int main(){ 50 scanf("%d", &n); 51 memset(cnt, -1, sizeof(cnt)); 52 for(int i = 1; i <= n; i++){ 53 scanf("%I64d", &A[i]); 54 mp[A[i]] = i; 55 } 56 for(int i = 1; i <= n; i++){ 57 if(cnt[i] == -1){ 58 dfs(A[i]); 59 } 60 61 } 62 while(1){ 63 int Max = 0, save; 64 for(int i = 1; i <= n; i++){ 65 if(cnt[i] >= Max && A[i] != 0) 66 Max = cnt[i], save = i; 67 } 68 if(Max == 0) break; 69 Show(A[save], Max); 70 } 71 return 0; 72 }
题意:求环的个数, 这个环需要这个环上的点都只有2条边。
题解:DFS一下, 从某个点出发, 如果按照一个方向走能回到该点, 并且路上的点都只有2条边, 就cnt++;
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define max3(a,b,c) max(a,max(b,c)) 12 #define min3(a,b,c) min(a,min(b,c)) 13 typedef pair<int,int> pll; 14 const int INF = 0x3f3f3f3f; 15 const LL mod = 1e9+7; 16 const int N = 1e5+10; 17 vector<int> son[2*N]; 18 int vis[N*2]; 19 int cnt = 0; 20 void dfs(int u, int l, int v){ 21 if(u == v){cnt++; return ;} 22 if(vis[v]) return ; 23 vis[v] = 1; 24 if(son[v].size() == 2){ 25 if(l == son[v][0]) dfs(u,v, son[v][1]); 26 else dfs(u,v,son[v][0]); 27 } 28 } 29 int main(){ 30 ///Fopen; 31 int n, m; 32 scanf("%d%d", &n, &m); 33 for(int i = 1; i <= m; i++){ 34 int u, v; 35 scanf("%d%d", &u, &v); 36 son[u].pb(v); 37 son[v].pb(u); 38 } 39 for(int i = 1; i <= n; i++){ 40 if(!vis[i]) { 41 vis[i] = 1; 42 if(son[i].size() == 2){ 43 dfs(i, i, son[i][0]); 44 } 45 } 46 } 47 printf("%d", cnt); 48 return 0; 49 }
题意:求最长的递增子序列。 这个序列的前一项与后一项相差为1。
题解:开一个map, 映射一下某个值上一次访问的位置,然后每次处理一个新的点, 如果比这个点小1的值的mp不为0, 就将这个点的位置指向上一个比他小1的点,记录长度, 如果没有比他小1的点, 就指向-1, 然后长度记为1, 然后每次更新最长的值 和 相应的位置就好了。
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define max3(a,b,c) max(a,max(b,c)) 12 #define min3(a,b,c) min(a,min(b,c)) 13 typedef pair<int,int> pll; 14 const int INF = 0x3f3f3f3f; 15 const LL mod = 1e9+7; 16 const int N = 2e5+10; 17 map<int,int> mp; 18 int cnt[N]; 19 int pre[N]; 20 int A[N]; 21 int Max = 0, B = 0; 22 void Show(int u){ 23 if(u == -1) return; 24 Show(pre[u]); 25 printf("%d ", u); 26 } 27 int main(){ 28 ///Fopen; 29 int n; 30 scanf("%d", &n); 31 for(int i = 1; i <= n; i++) 32 scanf("%d", &A[i]); 33 for(int i = 1; i <= n; i++){ 34 int t = A[i]-1; 35 if(mp.count(t)){ 36 pre[i] = mp[t]; 37 cnt[i] = cnt[pre[i]]+1; 38 } 39 else pre[i] = -1, cnt[i] = 1; 40 if(Max < cnt[i]){ 41 Max = cnt[i]; 42 B = i; 43 } 44 mp[A[i]] = i; 45 } 46 printf("%d\n", cnt[B]); 47 Show(B); 48 return 0; 49 }