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BUCT蓝桥杯热身赛I题解

时间:2014-11-23 14:21:55      阅读:264      评论:0      收藏:0      [点我收藏+]

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A 题 十六进制转换(hex)

看起来是一个水题,可是事实上确实是个水题。如果你按照题目的意思来,先把16进制转化成10进制,再把10进制转换成8进制,不好写。

所以从2进制的思想看16进制的数每一个数都是由4位二进制组成,然而8进制数是有3位二进制组成,所以我们大可把16进制先转成2进制,在从 二进制转成8进制,注意前导0.

注意姿势,由于姿势的问题,超时了3次,真冤。代码见下方

bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4  
 5 using namespace std;
 6  
 7 string const_ch[16] = {"0000","0001","0010","0011","0100","0101",
 8 "0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};
 9  
10 string const_cc[8] = {"000","001","010","011","100","101","110","111"};
11  
12 int Find(string str) {
13     for (int i = 0;i < 8;i++) {
14         if (str == const_cc[i]) return i;
15     }
16 }
17  
18 int main ( ) {
19     string str;
20     while (cin >> str) {
21         if (str == "0") cout << 0 << endl;
22         else {
23             string ans;
24             for (int i = 0,len = str.length();i < len;i ++) {
25                 ans += const_ch[str[i] >= A && str[i] <= F ? str[i] - A + 10 : str[i] - 0];
26             }
27             int len = ans.length();
28             if (len % 3 == 1){
29                 ans = "00" + ans;
30                 len += 2;
31             }
32             if (len % 3 == 2) {
33                 ans = "0" + ans;
34                 len ++;
35             }
36             int pos = 0;
37             int ANS[300000];
38             for (int i = 0;i < len;i += 3) {
39                 string temp = ans.substr(i,3);
40                 ANS[pos++] = Find(temp);
41             }
42             int temp = 0;
43             while (ANS[temp] == 0) temp++;
44             for (int i = temp;i < pos;i++) cout << ANS[i];
45                 cout << endl;
46         }
47     }
48 }
View Code

 

B题 聊斋(storys)

没什么好说的,直接暴力。

bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4  
 5 using namespace std;
 6  
 7 int n;
 8 string str;
 9 int mark[410];
10  
11 int extend1(int i,char ch,int dir) {
12     int pos = i;
13     int ans = 0;
14     while (!mark[pos] && (str[pos] == ch || str[pos] == w)) {
15         ans ++;
16         mark[pos] = 1;
17         if (ans >= n) return n;
18         pos = (pos + n + dir) % n;
19     }
20     return ans;
21 }
22  
23 int extend(int i,char ch,int dir) {
24     if (ch == w) {
25         return max(extend1(i,r,dir),extend1(i,b,dir));
26     }
27     else return extend1(i,ch,dir);
28 }
29  
30 int main () {
31     // freopen("1.in","r",stdin);
32     while (cin >> n) {
33         getchar();
34         cin >> str;
35         int maxn = -1;
36         for (int i = 0;i < n;i++) {
37             if (str[i] == w) {
38                 memset(mark,0,sizeof(mark));
39                 maxn = max(maxn,extend(i,r,-1) + extend((i + 1) % n,str[(i + 1) % n],1));
40                 memset(mark,0,sizeof(mark));
41                 maxn = max(maxn,extend(i,b,-1) + extend((i + 1) % n,str[(i + 1) % n],1));
42                 memset(mark,0,sizeof(mark));
43                 maxn = max(maxn,extend(i,r,1) + extend((i - 1 + n) % n,str[(i - 1 + n) % n],-1));
44                 memset(mark,0,sizeof(mark));
45                 maxn = max(maxn,extend(i,b,1) + extend((i - 1 + n) % n,str[(i - 1 + n) % n],-1));
46             }
47             else {
48                 memset(mark,0,sizeof(mark));
49                 maxn = max(maxn,extend(i,str[i],-1) + extend((i + 1) % n,str[(i + 1) % n],1));
50                 memset(mark,0,sizeof(mark));
51                 maxn = max(maxn,extend(i,str[i],1) + extend((i - 1 + n) % n,str[(i - 1 + n) % n],-1));
52             }
53         }
54         cout << maxn << endl;
55     }
56     return 0;
57 }
View Code

 

C题  奶牛的编号(privc)

刚看到这个题的时候,还以为是找逆序数,因为题目并没规定只能交换相邻的,所以就不是求逆序数了。后来想了想,因为题目之后1,2,3三个等级,所以我们可以

将所有的划分成3块,第一块放1,第二块放2,etc. 但是怎么交换才能保证交换次数最少呢,很简单,找到第一块中的其他数与相应的块里面的数做交换,比如在一块

中出现了2,那么我就找第二块中的1与之做交换。但是当交换完之后,会存在 这么一种情况不能被交换 2 3 1,所以最后要遍历一下看看有多少个这样的情况存在,

每种ans+=2;代码如下。

bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5  
 6 using namespace std;
 7  
 8 int main () {
 9     int a[1001];
10     int n;
11     int b[4] = {0};
12     // freopen("1.in","r",stdin);
13     cin >> n;
14     for (int i = 0;i < n;i++) {
15         cin >> a[i];
16         b[a[i]]++;
17     }
18     b[2] = b[1] + b[2];
19     b[3] = n;
20     int ans = 0;
21     for (int i = 1;i < 3;i++) { 
22         for (int j = b[i - 1];j < b[i];j++) {
23             if (a[j] == i) continue;
24             else {
25                 for (int k = b[a[j] - 1];k < b[a[j]];k++) {
26                     if (a[k] == i) {
27                         ans ++;
28                         swap(a[j],a[k]);
29                         break;
30                     }
31                 }
32             }
33         }
34     }
35     for (int i = 0;i < b[1];i++) {
36         if (a[i] == 1) continue;
37         else {
38             ans += 2;
39         }
40     }
41     cout << ans << endl;
42 }
View Code

 

下面是罗的代码

bubuko.com,布布扣
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define MAXN 1010
 6 using namespace std;
 7 int a[MAXN],c[MAXN],cnt;
 8 int main(){
 9     int n;
10     while(~scanf("%d",&n)){
11         for(int i=0;i<n;i++){
12             scanf("%d",&a[i]);
13             c[i]=a[i];
14         }
15         sort(c,c+n);
16         int ans=0;
17         for(int i=0;i<n;i++){
18             if(a[i]==c[i])    continue;
19             ans++;
20             int tmp=-1;
21             for(int j=i+1;j<n;j++){
22                 if(a[j]==c[i]){
23                     if(a[i]==c[j]){
24                         swap(a[j],a[i]);
25                         break;
26                     }
27                     else tmp=j;
28                 }
29             }
30             if(~tmp) swap(a[i],a[tmp]);
31         }
32         printf("%d\n",ans);
33     }
34 }
35  
View Code

 

 

D题  贝贝与国王

水题一道,求2的(n - 1)次幂 (下面是陈的代码,不想吐槽了。。。。)

bubuko.com,布布扣
 1 #include<iostream>
 2 using namespace std;
 3 string ss[101]={"1","2","4","8","16","32","64","128","256","512","1024","2048","4096","8192","16384","32768","65536","131072","262144","524288","1048576","2097152","4194304","8388608","16777216","33554432","67108864","134217728","268435456","536870912","1073741824","2147483648","4294967296","8589934592","17179869184","34359738368","68719476736","137438953472","274877906944","549755813888","1099511627776","2199023255552","4398046511104","8796093022208","17592186044416","35184372088832","70368744177664","140737488355328","281474976710656","562949953421312","1125899906842624","2251799813685248","4503599627370496","9007199254740992","18014398509481984","36028797018963968","72057594037927936","144115188075855872","288230376151711744","576460752303423488","1152921504606846976","2305843009213693952","4611686018427387904","9223372036854775808","18446744073709551616","36893488147419103232","73786976294838206464","147573952589676412928","295147905179352825856","590295810358705651712","1180591620717411303424","2361183241434822606848","4722366482869645213696","9444732965739290427392","18889465931478580854784","37778931862957161709568","75557863725914323419136","151115727451828646838272","302231454903657293676544","604462909807314587353088","1208925819614629174706176","2417851639229258349412352","4835703278458516698824704","9671406556917033397649408","19342813113834066795298816","38685626227668133590597632","77371252455336267181195264","154742504910672534362390528","309485009821345068724781056","618970019642690137449562112","1237940039285380274899124224","2475880078570760549798248448","4951760157141521099596496896","9903520314283042199192993792","19807040628566084398385987584","39614081257132168796771975168","79228162514264337593543950336","158456325028528675187087900672","316912650057057350374175801344","633825300114114700748351602688","1267650600228229401496703205376"};
 4 int main()
 5 {
 6     int n;
 7     while(cin>>n)
 8     {
 9         cout<<ss[n-1]<<endl;
10     }
11 }
View Code

 下面是张益宁的代码,可以学习以下,不过让他水过了,却是在意料之外哈

bubuko.com,布布扣
 1 #include<stdio.h>
 2 #include<math.h>
 3 int main()
 4 {
 5     int n;
 6     while(~scanf("%d",&n))
 7     {
 8         printf("%.lf",(double)pow(2,n-1));
 9     } 
10 } 
View Code

 

E题 未出现的子串(unapeared)

水题一道,题目让求没有出现的子串,那么我就将输入的字符串从头到尾分成几个集合,每个集合都包含[1..q](如果没有,也没办法)

所以出现一个这样的集合ans++;(为什么呢?如果这是一个不完整的集合,即没有出现所有的字母,那么就可以找没出现的字母,但是如果都出现了,我只能将长度

增加),代码如下。

bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4  
 5 using namespace std;
 6  
 7 int main ( ) {
 8     int n,q;
 9     // freopen("1.in","r",stdin);
10     while (cin >> n >> q) {
11         int ans = 1;
12         int a;
13         int temp = 0;
14         q = (1 << q) - 1;
15         for (int i = 0;i < n;i++) {
16             cin >> a;
17             a --;
18             temp |= (1 << a);
19             if (temp == q) {
20                 ans ++;
21                 temp = 0;
22             }
23         }
24         cout << ans << endl;
25     }
26 }
View Code

这个题也可以DP来搞,DP[k]表示以k开头的最小非子串长度,那么可以从末尾向头找,(其实思想还是一样的)下面是罗的代码

bubuko.com,布布扣
 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 int dp[10],A[100100];
 6 int main(){
 7     int n,q;
 8     while(~scanf("%d%d",&n,&q)){
 9         memset(dp,0,sizeof(dp));
10         for(int i=0;i<n;i++)
11             scanf("%d",A+i);
12         for(int i=n-1;i>=0;i--){
13             int tmp=99999999;
14             for(int j=1;j<=q;j++){
15                 tmp=min(tmp,dp[j]);
16             }
17             dp[A[i]]=tmp+1;
18         }
19         int ans=99999999;
20         for(int i=1;i<=q;i++)
21             ans=min(ans,dp[i]);
22         printf("%d\n",ans+1);
23     }
24 }
View Code

 

F题 棋子移动(chess)

特别恶心的模拟题,没意思,水题一道

bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 int n;
 8 
 9 void work(string str1,int cur) {
10     cout << "OOO*O**__*" + str1 << endl;
11     cout << "O__*O**OO*" + str1 << endl;
12     cout << "O*O*O*__O*" + str1 << endl;
13     cout << "__O*O*O*O*" + str1 << endl;
14     cout << "step=" << cur + 4 << endl;
15 }
16 
17 void dfs(string str,int n,int temp,int num,int cur) {
18     cout << str << endl;
19     if (n == 3) {
20         string str1;
21         for (int i = 0;i < num - 1;i++) {
22             str1 += "O*";
23         }
24         work(str1,cur);
25         return;
26     }else 
27     if (temp == 0) {
28         string str1;
29         string str2;
30         for (int i = 0;i < n;i++) {
31             str1 += O;
32             str2 += *;
33         }
34         string str3;
35         for (int i = 0;i < num;i++) {
36             str3 += "O*";
37         }
38         dfs(str1 + str2 + "__" + str3,n,!temp,num,cur + 1);
39     }else if (temp == 1) {
40         string str1;
41         string str2;
42         string str3;
43         for (int i = 0;i < n - 1;i++) {
44             str1 += O;
45             str2 += *;
46         }
47         num++;
48         for (int i = 0;i < num;i++) {
49             str3 += "O*";
50         }
51         dfs(str1 + "__" + str2 + str3,n - 1,!temp,num,cur + 1);
52     }
53 }
54 
55 int main () {
56     while (cin >> n) {
57         string str1;
58         string str2;
59         for (int i = 0;i < n - 1;i++) {
60             str1 += O;
61             str2 += *;
62         }
63         dfs(str1 + "__" + str2 + "O*",n - 1,0,1,1);
64     }
65 }
View Code

 

G题 字母游戏(letters)

水题一道,直接暴力,注意‘26’剪枝。

bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4  
 5 using namespace std;
 6  
 7 int dir[4][2] = {-1,0,0,1,1,0,0,-1};
 8 string str[30]; 
 9 int ans[26];
10 int n,m;
11 int maxn;
12 void dfs(int x,int y,int cur) {
13     if (cur > maxn) {
14         maxn = cur;
15     }
16     if (maxn == 26) return;
17     for (int i = 0;i < 4;i++) {
18         int xx = x + dir[i][0];
19         int yy = y + dir[i][1];
20         if (xx >= 0 && xx < n && yy >= 0 && yy < m && !ans[str[xx][yy] - A]) {
21             ans[str[xx][yy] - A] = 1;
22             dfs(xx,yy,cur + 1);
23             ans[str[xx][yy] - A] = 0;
24         }
25     }
26 }
27  
28 int main () {
29     // freopen("1.in","r",stdin);
30     while (cin >> n >> m) {
31         for (int i = 0;i < n;i++) {
32             cin >> str[i];
33         }
34         memset(ans,0,sizeof(ans));
35         maxn = -1;
36         ans[str[0][0] - A] = 1; 
37         dfs(0,0,1);
38         cout << maxn << endl;
39     }
40 }
View Code

 

BUCT蓝桥杯热身赛I题解

标签:style   blog   http   io   ar   color   os   sp   for   

原文地址:http://www.cnblogs.com/xiaoshanshan/p/4115199.html

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