【牛客練習賽13】 A B C D【康拓展開】 E【DP or 記憶化搜尋】 F 【思維】
阿新 • • 發佈：2019-01-01
A 幸運數字Ⅰ
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld
題目描述
定義一個數字為幸運數字當且僅當它的所有數位都是4或者7。
比如說，47、744、4都是幸運數字而5、17、467都不是。
現在，給定一個字串s，請求出一個字串，使得：
1、它所代表的整數是一個幸運數字；
2、它非空；
3、它作為s的子串（不是子序列）出現了最多的次數（不能為0次）。
請求出這個串（如果有多解，請輸出字典序最小的那一個）。
輸入描述:
串s（1 <= |s| <= 50)。s只包含數字字元，可以有前導零。
輸出描述:
一個串表示答案。
無解輸出-1。
示例1
輸入
047
輸出
4
示例2
輸入
16
輸出
-1
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned long long
const int N = 3e4;
const int M = 10000+1;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
const LL inff = 0x3f3f3f3f3f3f3f3f;
int cnt[N];
int main(){
string B 幸運數字Ⅱ
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld
題目描述
定義一個數字為幸運數字當且僅當它的所有數位都是4或者7。
比如說，47、744、4都是幸運數字而5、17、467都不是。
定義next(x)為大於等於x的第一個幸運數字。給定l，r，請求出next(l) + next(l + 1) + … + next(r - 1) + next(r)。
輸入描述:
兩個整數l和r (1 <= l <= r <= 1000,000,000)。
輸出描述:
一個數字表示答案。
示例1
輸入
2 7
輸出
33
示例2
輸入
7 7
輸出
7
分析： 超暴力 解法，直接暴力求出從1到第一個大於1e9中的幸運數字
1000多點。
注意第一個大於1e9的幸運數字，程式暴力跑不出來，所以我們可以找一下規律 從而得到第一個大於1e9的幸運數字
程式碼
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned long long
const int N = 100;
const int M = 10000+1;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
const LL inff = 0x3f3f3f3f3f3f3f3f;
LL X[10000]={0,4,7,44,47,74,77,444,447,474,477,744,747,774,777,4444,4447,4474,4477,4744,4747,4774,4777,7444,7447,7474,7477,7744,7747,7774,7777,44444,44447,44474,44477,44744,44747,44774,44777,47444,47447,47474,47477,47744,47747,47774,47777,74444,74447,74474,74477,74744,74747,74774,74777,77444,77447,77474,77477,77744,77747,77774,77777,444444,444447,444474,444477,444744,444747,444774,444777,447444,447447,447474,447477,447744,447747,447774,447777,474444,474447,474474,474477,474744,474747,474774,474777,477444,477447,477474,477477,477744,477747,477774,477777,744444,744447,744474,744477,744744,744747,744774,744777,747444,747447,747474,747477,747744,747747,747774,747777,774444,774447,774474,774477,774744,774747,774774,774777,777444,777447,777474,777477,777744,777747,777774,777777,4444444,4444447,4444474,4444477,4444744,4444747,4444774,4444777,4447444,4447447,4447474,4447477,4447744,4447747,4447774,4447777,4474444,4474447,4474474,4474477,4474744,4474747,4474774,4474777,4477444,4477447,4477474,4477477,4477744,4477747,4477774,4477777,4744444,4744447,4744474,4744477,4744744,4744747,4744774,4744777,4747444,4747447,4747474,4747477,4747744,4747747,4747774,4747777,4774444,4774447,4774474,4774477,4774744,4774747,4774774,4774777,4777444,4777447,4777474,4777477,4777744,4777747,4777774,4777777,7444444,7444447,7444474,7444477,7444744,7444747,7444774,7444777,7447444,7447447,7447474,7447477,7447744,7447747,7447774,7447777,7474444,7474447,7474474,7474477,7474744,7474747,7474774,7474777,7477444,7477447,7477474,7477477,7477744,7477747,7477774,7477777,7744444,7744447,7744474,7744477,7744744,7744747,7744774,7744777,7747444,7747447,7747474,7747477,7747744,7747747,7747774,7747777,7774444,7774447,7774474,7774477,7774744,7774747,7774774,7774777,7777444,7777447,7777474,7777477,7777744,7777747,7777774,7777777,44444444,44444447,44444474,44444477,44444744,44444747,44444774,44444777,44447444,44447447,44447474,44447477,44447744,44447747,44447774,44447777,44474444,44474447,44474474,44474477,44474744,44474747,44474774,44474777,44477444,44477447,44477474,44477477,44477744,44477747,44477774,44477777,44744444,44744447,44744474,44744477,44744744,44744747,44744774,44744777,44747444,44747447,44747474,44747477,44747744,44747747,44747774,44747777,44774444,44774447,44774474,44774477,44774744,44774747,44774774,44774777,44777444,44777447,44777474,44777477,44777744,44777747,44777774,44777777,47444444,47444447,47444474,47444477,47444744,47444747,47444774,47444777,47447444,47447447,47447474,47447477,47447744,47447747,47447774,47447777,47474444,47474447,47474474,47474477,47474744,47474747,47474774,47474777,47477444,47477447,47477474,47477477,47477744,47477747,47477774,47477777,47744444,47744447,47744474,47744477,47744744,47744747,47744774,47744777,47747444,47747447,47747474,47747477,47747744,47747747,47747774,47747777,47774444,47774447,47774474,47774477,47774744,47774747,47774774,47774777,47777444,47777447,47777474,47777477,47777744,47777747,47777774,47777777,74444444,74444447,74444474,74444477,74444744,74444747,74444774,74444777,74447444,74447447,74447474,74447477,74447744,74447747,74447774,74447777,74474444,74474447,74474474,74474477,74474744,74474747,74474774,74474777,74477444,74477447,74477474,74477477,74477744,74477747,74477774,74477777,74744444,74744447,74744474,74744477,74744744,74744747,74744774,74744777,74747444,74747447,74747474,74747477,74747744,74747747,74747774,74747777,74774444,74774447,74774474,74774477,74774744,74774747,74774774,74774777,74777444,74777447,74777474,74777477,74777744,74777747,74777774,74777777,77444444,77444447,77444474,77444477,77444744,77444747,77444774,77444777,77447444,77447447,77447474,77447477,77447744,77447747,77447774,77447777,77474444,77474447,77474474,77474477,77474744,77474747,77474774,77474777,77477444,77477447,77477474,77477477,77477744,77477747,77477774,77477777,77744444,77744447,77744474,77744477,77744744,77744747,77744774,77744777,77747444,77747447,77747474,77747477,77747744,77747747,77747774,77747777,77774444,77774447,77774474,77774477,77774744,77774747,77774774,77774777,77777444,77777447,77777474,77777477,77777744,77777747,77777774,77777777,444444444,444444447,444444474,444444477,444444744,444444747,444444774,444444777,444447444,444447447,444447474,444447477,444447744,444447747,444447774,444447777,444474444,444474447,444474474,444474477,444474744,444474747,444474774,444474777,444477444,444477447,444477474,444477477,444477744,444477747,444477774,444477777,444744444,444744447,444744474,444744477,444744744,444744747,444744774,444744777,444747444,444747447,444747474,444747477,444747744,444747747,444747774,444747777,444774444,444774447,444774474,444774477,444774744,444774747,444774774,444774777,444777444,444777447,444777474,444777477,444777744,444777747,444777774,444777777,447444444,447444447,447444474,447444477,447444744,447444747,447444774,447444777,447447444,447447447,447447474,447447477,447447744,447447747,447447774,447447777,447474444,447474447,447474474,447474477,447474744,447474747,447474774,447474777,447477444,447477447,447477474,447477477,447477744,447477747,447477774,447477777,447744444,447744447,447744474,447744477,447744744,447744747,447744774,447744777,447747444,447747447,447747474,447747477,447747744,447747747,447747774,447747777,447774444,447774447,447774474,447774477,447774744,447774747,447774774,447774777,447777444,447777447,447777474,447777477,447777744,447777747,447777774,447777777,474444444,474444447,474444474,474444477,474444744,474444747,474444774,474444777,474447444,474447447,474447474,474447477,474447744,474447747,474447774,474447777,474474444,474474447,474474474,474474477,474474744,474474747,474474774,474474777,474477444,474477447,474477474,474477477,474477744,474477747,474477774,474477777,474744444,474744447,474744474,474744477,474744744,474744747,474744774,474744777,474747444,474747447,474747474,474747477,474747744,474747747,474747774,474747777,474774444,474774447,474774474,474774477,474774744,474774747,474774774,474774777,474777444,474777447,474777474,474777477,474777744,474777747,474777774,474777777,477444444,477444447,477444474,477444477,477444744,477444747,477444774,477444777,477447444,477447447,477447474,477447477,477447744,477447747,477447774,477447777,477474444,477474447,477474474,477474477,477474744,477474747,477474774,477474777,477477444,477477447,477477474,477477477,477477744,477477747,477477774,477477777,477744444,477744447,477744474,477744477,477744744,477744747,477744774,477744777,477747444,477747447,477747474,477747477,477747744,477747747,477747774,477747777,477774444,477774447,477774474,477774477,477774744,477774747,477774774,477774777,477777444,477777447,477777474,477777477,477777744,477777747,477777774,477777777,744444444,744444447,744444474,744444477,744444744,744444747,744444774,744444777,744447444,744447447,744447474,744447477,744447744,744447747,744447774,744447777,744474444,744474447,744474474,744474477,744474744,744474747,744474774,744474777,744477444,744477447,744477474,744477477,744477744,744477747,744477774,744477777,744744444,744744447,744744474,744744477,744744744,744744747,744744774,744744777,744747444,744747447,744747474,744747477,744747744,744747747,744747774,744747777,744774444,744774447,744774474,744774477,744774744,744774747,744774774,744774777,744777444,744777447,744777474,744777477,744777744,744777747,744777774,744777777,747444444,747444447,747444474,747444477,747444744,747444747,747444774,747444777,747447444,747447447,747447474,747447477,747447744,747447747,747447774,747447777,747474444,747474447,747474474,747474477,747474744,747474747,747474774,747474777,747477444,747477447,747477474,747477477,747477744,747477747,747477774,747477777,747744444,747744447,747744474,747744477,747744744,747744747,747744774,747744777,747747444,747747447,747747474,747747477,747747744,747747747,747747774,747747777,747774444,747774447,747774474,747774477,747774744,747774747,747774774,747774777,747777444,747777447,747777474,747777477,747777744,747777747,747777774,747777777,774444444,774444447,774444474,774444477,774444744,774444747,774444774,774444777,774447444,774447447,774447474,774447477,774447744,774447747,774447774,774447777,774474444,774474447,774474474,774474477,774474744,774474747,774474774,774474777,774477444,774477447,774477474,774477477,774477744,774477747,774477774,774477777,774744444,774744447,774744474,774744477,774744744,774744747,774744774,774744777,774747444,774747447,774747474,774747477,774747744,774747747,774747774,774747777,774774444,774774447,774774474,774774477,774774744,774774747,774774774,774774777,774777444,774777447,774777474,774777477,774777744,774777747,774777774,774777777,777444444,777444447,777444474,777444477,777444744,777444747,777444774,777444777,777447444,777447447,777447474,777447477,777447744,777447747,777447774,777447777,777474444,777474447,777474474,777474477,777474744,777474747,777474774,777474777,777477444,777477447,777477474,777477477,777477744,777477747,777477774,777477777,777744444,777744447,777744474,777744477,777744744,777744747,777744774,777744777,777747444,777747447,777747474,777747477,777747744,777747747,777747774,777747777,777774444,777774447,777774474,777774477,777774744,777774747,777774774,777774777,777777444,777777447,777777474,777777477,777777744,777777747,777777774,777777777,4444444444};
int sz=1024;
LL sum[10000];
int main(){
for(int i=1;i<sz;i++) { sum[i]=sum[i-1]+X[i]; }
int l,r;cin>>l>>r;
int a=lower_bound(X,X+sz,l)-X;
LL ans=0;int i; int L=l;
for( i=a;i<sz;i++){
if(X[i]>=r){
ans+=(r-L+1)*X[i];
break;
}
ans+=X[i]*(X[i]-L+1);
L=X[i]+1;
}
printf("%lld\n",ans);
return 0;
}
分析二 ：因為我們知道數字只有4和7，所以我們可以通過二進位制列舉 （或者bfs，dfs）得到長度小於等於10的所有情況，然後sort一下。
C 幸運數字Ⅲ
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld
題目描述
定義一個數字為幸運數字當且僅當它的所有數位都是4或者7。
比如說，47、744、4都是幸運數字而5、17、467都不是。
假設現在有一個數字d，現在想在d上重複k次操作。
假設d有n位，用d1,d2,…,dn表示。
對於每次操作，我們想要找到最小的x (x < n)，使得dx=4並且dx+1=7。
如果x為奇數，那麼我們把dx和dx+1都變成4；
否則，如果x為偶數，我們把dx和dx+1都變成7；
如果不存在x，那麼我們不做任何修改。
現在請問k次操作以後，d會變成什麼樣子。
輸入描述:
第一行兩個整數n，k表示d的長度和操作次數。
第二行一個數表示d。資料保證不存在前導零。
1 <= n <= 100,000
0 <= k <= 1000,000,000
輸出描述:
一個數字表示答案。
示例1
輸入
7 4
4727447
輸出
4427477
示例2
輸入
4 2
4478
輸出
4478
分析： 可以發現477或者447這種的並且4所在位置為奇數時候，會有迴圈。而其他情況下，都是正想不可逆的，所以遇到一個447或者477並且4位置為奇數的情況就可以停止，然後根據剩下次數的奇偶來決定最後序列為什麼。
程式碼
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned long long
const int N = 1e5+11;
const int M = 10000+1;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
const LL inff = 0x3f3f3f3f3f3f3f3f;
char s[N];
int main(){
LL n,k;cin>>n>>k;
scanf("%s",s+1);
LL cnt=0;
for(int i=1;i<=n;i++){
if(s[i]=='4' &&i+1<=n && s[i+1]=='7' &&i+2<=n && s[i+2]=='7') {
if(i&1){
LL cha=k-cnt;
if(cha&1) {
s[i+1]='4';
for(int i=1;i<=n;i++) printf("%c",s[i]);
return 0;
}else{
for(int i=1;i<=n;i++) printf("%c",s[i]);
return 0;
}
}
}
if(s[i]=='4' && i+1<=n && s[i+1]=='7') {
if(i&1) {
s[i]=s[i+1]='4';
}else {
s[i]=s[i+1]='7';
}
cnt++;
if(cnt==k) break;
if(i>=2) i-=2;
else if(i>=1) i-=1;
}
}
for(int i=1;i<=n;i++) printf("%c",s[i]);
return 0;
}
D 幸運數字Ⅳ
E 烏龜跑步
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld
題目描述
有一隻烏龜，初始在0的位置向右跑。
這隻烏龜會依次接到一串指令，指令T表示向後轉，指令F表示向前移動一個單位。烏龜不能忽視任何指令。
現在我們要修改其中正好n個指令（一個指令可以被改多次，一次修改定義為把某一個T變成F或把某一個F變成T）。
求這隻烏龜在結束的時候離起點的最遠距離。（假設烏龜最後的位置為x，我們想要abs(x)最大，輸出最大的abs(x)）
輸入描述:
第一行一個字串c表示指令串。c只由F和T構成。
第二行一個整數n。
1 <= |c| <= 100, 1 <= n <= 50
輸出描述:
一個數字表示答案。
示例1
輸入
FT
1
輸出
2
示例2
輸入
FFFTFFF
2
輸出
6
分析： 一看題目 就知道 是DP方向的，DP我基本不會，但是我會記憶化搜尋，對於一部分DP來說，只要用DFS可以遍歷其所有狀態，我加上記憶化，那麼一定可以過【狀態一定要能夠用陣列來儲存】。
比如這道題 ： dfs(int now,int opc,int sta,int pre) 表示 現在該訪問now這個命令，opc表示當前還剩下幾個可改變的操作，sta表示當前是正向還是反向，pre表示我是站在那個位置來執行當前的命令。
可以發現任意時間我都可以用上述 幾個引數 都可以表示當前狀態，這個就是關鍵——-是否可以將DP轉化為記憶化搜尋。
程式碼寫的有點醜了，但是隻要知道上述的思路，記憶化搜尋（暴力）應該很好寫吧。
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ULL unsigned long long
const int N = 200+11;
const int M = 110;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
const LL inff = 0x3f3f3f3f3f3f3f3f;
char s[N]; int len;
int dp[N][N][2][N*2];
int dfs(int now,int opc,int sta,int pre){
if(now==len+1 && !opc) return pre;
if(now==len+1) return M;
if(dp[now][opc][sta][pre]!=-1) return dp[now][opc][sta][pre];
int ans=0,cha=0;
if(s[now]=='F') {
if(sta){
int a=dfs(now+1,opc,sta,pre+1 ) ;
int b=abs(a-M);
if( b>=abs(cha)){
cha=b;
ans=a;
}
}else {
int a=dfs(now+1,opc,sta,pre-1) ;
int b=abs(a-M);
if( b>=abs(cha)){
cha=b;
ans=a;
}
}
}else {
if(sta){
int a=dfs(now+1,opc,sta^1,pre ) ;
int b=abs(a-M);
if( b>=abs(cha)){
cha=b;
ans=a;
}
}else {
int a=dfs(now+1,opc,sta^1,pre) ;
int b=abs(a-M);
if( b>=abs(cha)){
cha=b;
ans=a;
}
}
}
if(opc>0){
if(s[now]=='T') {
if(sta){
int a=dfs(now+1,opc-1,sta,pre+1 ) ;
int b=abs(a-M);
if( b>=abs(cha)){
cha=b;
ans=a;
}
}else{
int a=dfs(now+1,opc-1,sta,pre-1) ;
int b=abs(a-M);
if( b>=abs(cha)){
cha=b;
ans=a;
}
}
}else {
if(sta){
int a=dfs(now+1,opc-1,sta^1,pre ) ;
int b=abs(a-M);
if( b>=abs(cha)){
cha=b;
ans=a;
}
}else {
int a=dfs(now+1,opc-1,sta^1,pre) ;
int b=abs(a-M);
if( b>=abs(cha)){
cha=b;
ans=a;
}
}
}
}
return dp[now][opc][sta][pre]=ans;
}
int main(){
memset(dp,-1,sizeof(dp));
int n; scanf("%s %d",s+1,&n); len=strlen(s+1);
dfs(1,n,1,M); // 從M點開始，這樣才不會導致陣列越界 ，因為dfs中pre表示當前位置
cout<<abs(dp[1][n][1][M]-M);
return 0;
}