北郵2010網(wǎng)研院和計(jì)算機(jī)院復(fù)試上機(jī)題及代碼

1、王道（）友情！予人玫瑰 手留余香！2010 北郵網(wǎng)研院復(fù)試上機(jī)題目：第一題：查找 輸入數(shù)組長(zhǎng)度 n輸入數(shù)組a1.n輸入查找個(gè)數(shù)m輸入查找數(shù)字b1.m輸出 YES or NO 查找有則 YES否則 NO如(括號(hào)內(nèi)容為注釋) 輸入：5(數(shù)組長(zhǎng)度)1 5 2 4 3(數(shù)組)3(查找個(gè)數(shù))2 5 6(查找具體數(shù)字)輸出： YES YESNO#include <stdio.h>#include <stdlib.h>int main()int n,m,i,j,a2001=0,b2001=0,flag=0; scanf("%d",&n);for(i=0;i

2、<n;i+) scanf("%d",&ai);scanf("%d",&m); for(i=0;i<m;i+) scanf("%d",&bi); for(i=0;i<m;i+)for(j=0;j<n;j+)if(bi=aj)flag=1; printf("YESn"); break;王道王道（）友情！予人玫瑰手留余香！if(flag!=1) printf("NOn"); flag=0;/ / system("PAUSE"); re

3、turn 0;第二題：查找第K 小數(shù)查找一個(gè)數(shù)組的第K 小的數(shù)，注意同樣大小算一樣大如 2 1 3 4 5 2 第三小數(shù)為 3如(括號(hào)內(nèi)容為注釋) 輸入：6(數(shù)組長(zhǎng)度 n)2 1 3 5 2 2(數(shù)組)3(K 即為第三小數(shù))輸出：3Code#include <stdio.h> #include <stdlib.h>int main(int argc, char *argv)int n,k,i,j,a1001,temp,m=1;scanf("%d %d",&n,&k); for(i=0;i<n;i+) scanf("%d

4、",&ai);for(i=0;i<n-1;i+) for(j=0;j<n-i-1;j+)if(aj>aj+1)temp=aj; aj=aj+1; aj+1=temp;for(i=0;i<n-1;i+)if(ai=ai+1)&&(i<k)王道王道（）友情！予人玫瑰 手留余香！k+;printf("%dn",ak-1);/ system("PAUSE");return 0;·第三題：牌只有 1 到 9，手里拿著已經(jīng)排好序的牌 a，對(duì)方出牌 b，用程序夠壓過對(duì)方出牌規(guī)則：出牌牌型有 5

5、種手中牌是否能1 一張2 兩張3 三張4 四張5 五張如 4 則 5.9 可壓過如 44 則 55，66，77，.，99 可壓過如 444 規(guī)則如2如 4444 規(guī)則如2牌型只有 12345 23456 34567 45678 56789 五個(gè)，后面的比前面的均大壓過輸出YES 否則NO如(括號(hào)內(nèi)容為注釋) 輸入：12233445566677(手中牌)33(出牌)輸出：YESProblem Id: 1820Submit time: 2010-04-25 20:28:36 User_id: jyjyjy1989Memory:204KTime:19MSLanguage:G+ Result:Acce

6、pted··Code#include <iostream> #include<string.h>using namespace std;····int main(void)王道王道（）友情！予人玫瑰手留余香！·························

7、3;··char str1100,str25; scanf("%s",str1);char ch;ch=getchar(); scanf("%s",str2);int i=0;int count10=0; while(str1i!='n')if(str1i-'1')=0) count0+;else else else else else else else else else i+;if(str1i-'1')=1)if(str1i-'1')=2)if(str1i-'

8、;1')=3)if(str1i-'1')=4)count1+; count2+; count3+;count4+;if (str1i-'1')=5)count5+;if(str1i-'1')=6)if(str1i-'1')=7)if(str1i-'1')=8) break;count6+;count7+; count8+;ints1,s2,s3,s4,s5;int flag=1;if(strlen(str2)=1)s1=*str2-'1'for(;s1<9;s1+) if(counts

9、1+1>0)printf("YESn",s1);flag=0;break;else if(strlen(str2)=2) s2=*str2-'1'>=2) printf("YESn",s2);flag=0;break; else if(strlen(str2)=3) s3=*str2-'1'>=3) printf("YESn");flag=0;break; else if(strlen(str2)=4) s4=*str2-'1'>=4) printf("

10、YESn");flag=0;break; for(;s2<9;s2+)if(counts2+1·for(;s3<9;s3+)if(counts3+1·for(;s4<9;s4+)if(counts4+1·else if(strlen(str2)=5)s5=*str2-'1' for(;s5<9;s5+) if(counts5+5>0&&counts5+1>0&&counts5+2>0&&counts5+3>0&&counts5+

11、4>0&&(s5+5)< 9) printf("YESn");flag=0;break;if(flag=1) printf("NOn");/ system("PAUSE"); return EXIT_SUCCESS;····第四題：樹 查找簡(jiǎn)單說就是一棵樹，輸出某一深度的所有節(jié)點(diǎn)，有則輸出這些節(jié)點(diǎn)，無則輸出 EMPTY，具王道王道（）友情！予人玫瑰 手留余香！體描述得借助圖形比較好，懶得寫了，基本就是這個(gè)樣子的。2010 北郵計(jì)算機(jī)學(xué)院上機(jī)題目回憶版題目大意（回憶版）

12、：第一行輸入一個(gè)數(shù)，為 n，第二行輸入 n 個(gè)數(shù)，這 n 個(gè)數(shù)中，如果偶數(shù)比奇數(shù)多，輸出 NO，否則輸出 YES。Sample:Input:51 2 3 4 5Output:YESProblem Id: 1814Submit time: 2010-04-10 14:06:48 User_id: bupt111310352 Memory:72KTime:17MS Language:G+Result:AcceptedCode*/#include <stdio.h>int main()/ 所有數(shù)的數(shù)目int m_Num_Count; scanf("%d", &

13、m_Num_Count);/ 輸入數(shù)int m_Input_Num;/ 偶數(shù) 奇數(shù)的個(gè)數(shù)int m_Oushu_Count = 0, m_Jishu_Count = 0;/ 循環(huán)輸入王道王道（）友情！予人玫瑰手留余香！while(m_Num_Count > 0)scanf("%d", &m_Input_Num); getchar();/奇偶并計(jì)數(shù)if( (m_Input_Num % 2) = 0 )+ m_Oushu_Count;else+ m_Jishu_Count;- m_Num_Count;/ 輸出if(m_Jishu_Count >= m_Ous

14、hu_Count) printf("YESn");elseprintf("NOn");return 0;/*題目大意（回憶版）：第一行輸入一個(gè)數(shù) n，1 <= n <= 1000，下面輸入n 行數(shù)據(jù)，每一行有兩個(gè)數(shù)，分別是 x y。輸出一組 x y，該組數(shù)據(jù)是所有數(shù)據(jù)中 x 最小，且在x 相等的情況下 y 最小的。Sample:Input: 53 32 25 52 13 6Output:2 1Problem Id: 1815Submit time: 2010-04-10 14:24:39 User_id: bupt111310352王道王道（

15、）友情！予人玫瑰手留余香！Memory:104KLanguage:G+Time:19MSResult:AcceptedCode*/#include <stdio.h>typedef struct LNodeint m_X; int m_Y;bool m_Used;/ x 坐標(biāo)值/ y 坐標(biāo)值/ 該位是否被使用;int main()/ 樣例個(gè)數(shù)int m_Case_Count;scanf( "%d", &m_Case_Count );/ 因?yàn)?1 <= N <= 1000，建結(jié)點(diǎn)LNode m_Case 1000 ;/ 循環(huán)數(shù)int m_Cyc

16、le;/ 最小的 x, y 值，并初始化，1 <= x,y <= 1000 int m_X_Min = 1000, m_Y_Min = 1000;/ 初始化結(jié)點(diǎn)被使用狀態(tài)for( m_Cycle = 0; m_Cycle < 1000; + m_Cycle ) m_Case m_Cycle .m_Used = false;/ 初始化循環(huán)數(shù)，下面使用m_Cycle = 0;/ 循環(huán)輸入樣例while( m_Case_Count > 0 )/ 輸入坐標(biāo)值scanf( "%d %d", &m_Case m_Cycle .m_X, &m_Ca

17、se m_Cycle .m_Y );/ 置狀態(tài)值，并且 m_Cycle +王道王道（）友情！予人玫瑰 手留余香！m_Case m_Cycle + .m_Used = true;-m_Case_Count;/ 循環(huán)找最小的x,y 值for( m_Cycle = 0; ( m_Cycle < 1000 ) && ( m_Case m_Cycle .m_Used ) ; + m_Cycle )if( m_Case m_Cycle .m_X< m_X_Min )/ 如果找到更小的 x 值， m_X_Min = m_Case m_Cycle .m_X; m_Y_Min = m

18、_Case m_Cycle .m_Y;else if ( m_Case m_Cycle .m_X= m_X_Min )/ 如果 x 值相等y 值, 如果在相等的 x 值的條件下，找到更小的的 y 值，yif( m_Case m_Cycle .m_Y< m_Y_Min )m_Y_Min = m_Case m_Cycle .m_Y;/ 輸出找到的最小值printf("%d %dn", m_X_Min, m_Y_Min );return 0;/*題目大意（回憶版）：該題是要翻轉(zhuǎn)數(shù)據(jù)。首先輸入一個(gè) 5 * 5 的數(shù)組，然后輸入一行，這一行有四個(gè)數(shù)，前兩個(gè)代表操作類型，后兩個(gè)數(shù)

19、 x y 代表需操作數(shù)據(jù)為以 x y 為左上角的個(gè)數(shù)據(jù)。操作類型有四種：1 21 32 22 3表示：90 度，順時(shí)針，翻轉(zhuǎn) 4 個(gè)數(shù)表示：90 度，順時(shí)針，翻轉(zhuǎn) 9 個(gè)數(shù)表示：90 度，逆時(shí)針，翻轉(zhuǎn) 4 個(gè)數(shù)表示：90 度，逆時(shí)針，翻轉(zhuǎn) 9 個(gè)數(shù)Sample:王道王道（）友情！予人玫瑰手留余香！Input:1 2 3 4 56 7 8 9 1011 12 13 14 1516 17 18 19 2021 22 23 24 251 3 1 1Output:11 6 1 4 512 7 2 9 1013 8 3 14 1516 17 18 19 2021 22 23 24 25Problem I

20、d: 1816Submit time: 2010-04-10 15:06:04 User_id: bupt111310352 Memory:88KTime:14MS Language:G+Result:Accepted解題思路：- -。細(xì)心點(diǎn)就行。Code*/#include <stdio.h>int main()/ 輸入數(shù)據(jù)int m_Case 5 5 ;scanf( "%d %d %d %d %d", &m_Case 0 0 , &m_Case 0 1 , &m_Case 0 2 , &m_Case 0 3 , &m_

21、Case 0 4 );scanf( "%d %d %d %d %d", &m_Case 1 0 , &m_Case 1 1 , &m_Case 1 2 , &m_Case 1 3 , &m_Case 1 4 );scanf( "%d %d %d %d %d", &m_Case 2 0 , &m_Case 2 1 , &m_Case 2 2 , &m_Case 2 3 , &m_Case 2 4 );scanf( "%d %d %d %d %d", &

22、;m_Case 3 0 , &m_Case 3 1 , &m_Case 3 2 , &m_Case 3 3 , &m_Case 3 4 );scanf( "%d %d %d %d %d", &m_Case 4 0 , &m_Case 4 1 , &m_Case 4 2 ,&m_Case 4 3 , &m_Case 4 4 );/ 輸入操作王道王道（）友情！予人玫瑰 手留余香！int m_Op_First, m_Op_Second, m_X, m_Y;scanf( "%d %d %d %d&qu

23、ot;, &m_Op_First, &m_Op_Second, &m_X, &m_Y );/ 臨時(shí)數(shù)據(jù)int m_Temp_First, m_Temp_Second;if( ( m_Op_First = 1 ) && ( m_Op_Second = 2 ) )/ 如果是 1 2, 翻轉(zhuǎn) 4 個(gè)數(shù)，90 度，順時(shí)針m_Temp_First = m_Case m_X - 1 m_Y - 1 ;m_Case m_X - 1 m_Y - 1 = m_Case m_X m_Y - 1 ; m_Case m_X m_Y - 1 = m_Case m_X m_

24、Y ; m_Case m_X m_Y = m_Case m_X - 1 m_Y ; m_Case m_X - 1 m_Y = m_Temp_First;else if( ( m_Op_First = 1 ) && ( m_Op_Second = 3 ) )/ 如果是 1 3，翻轉(zhuǎn) 9 個(gè)數(shù)，90 度，順時(shí)針m_Temp_First = m_Case m_X - 1 m_Y - 1 ; / 00m_Temp_Second = m_Case m_X m_Y - 1 ;/ 10m_Case m_X - 1 m_Y - 1 = m_Case m_X + 1 m_Y - 1 ; / 00

25、 = 20m_Case m_X m_Y - 1 = m_Case m_X + 1 m_Y ; m_Case m_X + 1 m_Y - 1 = m_Case m_X + 1 m_Y + 1 ; m_Case m_X + 1 m_Y = m_Case m_X m_Y + 1 ; m_Case m_X + 1 m_Y + 1 = m_Case m_X - 1 m_Y + 1 ; m_Case m_X m_Y + 1 = m_Case m_X - 1 m_Y ; m_Case m_X - 1 m_Y + 1 = m_Temp_First;m_Case m_X - 1 m_Y = m_Temp_Sec

26、ond;/ 10 = 21/ 20 = 22/ 21 = 12/ 22 = 02/ 12 = 01/ 02 = 00/ 01 = 10else if( ( m_Op_First = 2 ) && ( m_Op_Second = 2 ) )/ 如果是 2 2, 翻轉(zhuǎn) 4 個(gè)數(shù)，90 度，逆時(shí)針m_Temp_First = m_Case m_X - 1 m_Y - 1 ;m_Case m_X - 1 m_Y - 1 = m_Case m_X - 1 m_Y ; m_Case m_X - 1 m_Y = m_Case m_X m_Y ; m_Case m_X m_Y = m_Case

27、 m_X m_Y - 1 ; m_Case m_X m_Y - 1 = m_Temp_First;else if( ( m_Op_First = 2 ) && ( m_Op_Second = 3 ) )/ / 如果是 2 3，翻轉(zhuǎn) 9 個(gè)數(shù)，90 度，逆時(shí)針m_Temp_First = m_Case m_X - 1 m_Y - 1 ; m_Temp_Second = m_Case m_X - 1 m_Y ;/ 00/ 01m_Case m_X - 1 m_Y - 1 = m_Case m_X - 1 m_Y + 1 ; / 00 = 02王道王道（）友情！予人玫瑰 手留余香！m

28、_Case m_X - 1 m_Y = m_Case m_X m_Y + 1 ;/ 01 = 12m_Case m_X - 1 m_Y + 1 = m_Case m_X + 1 m_Y + 1 ; / 02 = 22m_Case m_X m_Y + 1 = m_Case m_X + 1 m_Y ;/ 12 = 21m_Case m_X + 1 m_Y + 1 = m_Case m_X + 1 m_Y - 1 ; / 22 = 20m_Case m_X + 1 m_Y = m_Case m_X m_Y - 1 ; m_Case m_X + 1 m_Y - 1 = m_Temp_First;m_C

29、ase m_X m_Y - 1 = m_Temp_Second;/ 21 = 10/ 20 = 00/ 10 = 01/ 輸出printf( "%d %d %d %d %dn", m_Case 0 0 , m_Case 0 m_Case 0 3 , m_Case 0 4 );1 , m_Case 0 2 ,printf( "%d %d %d %d %dn", m_Case 1 0 , m_Case 1 1 , m_Case 1 2 , m_Case 1 3 , m_Case 1 4 );printf( "%d %d %d %d %dn"

30、;, m_Case 2 0 , m_Case 2 1 , m_Case 2 2 , m_Case 2 3 , m_Case 2 4 );printf( "%d %d %d %d %dn", m_Case 3 0 , m_Case 3 1 , m_Case 3 2 , m_Case 3 3 , m_Case 3 4 );printf( "%d %d %d %d %dn", m_Case 4 0 , m_Case 4 1 , m_Case 4 2 ,m_Case 4 3 , m_Case 4 4 );return 0;/*/ 我最糾結(jié)的題。前 3 個(gè)題一個(gè)小

31、時(shí)。這個(gè)題一個(gè)小時(shí)都沒有弄完。我實(shí)在是很郁悶啊。當(dāng)時(shí)真感覺想法太多了。還不如好好分析分析。這個(gè)題主要是要設(shè) parent 指針。如果這題還需要輸出編碼的話，則需要同時(shí)設(shè) rchild 與 lchild 指針。太糾結(jié)了這道題。這都沒做出來，強(qiáng)烈bs?；丶遗藭?huì)就弄出來了。樹，第一行輸入一個(gè)數(shù) n，表示葉結(jié)點(diǎn)的個(gè)數(shù)。需要用這些葉題目大意（回憶版）：結(jié)點(diǎn)生成結(jié)點(diǎn)的值與樹，根據(jù)的乘積之和。樹的概念，這些結(jié)點(diǎn)值，即 weight，題目需要輸出所有Sample:Input:51 2 2 5 9Output:37樹，結(jié)點(diǎn) 1 的為 4，結(jié)點(diǎn) 2 的為 4，結(jié)點(diǎn) 2 的為 3，解釋：即生成如下圖王道王道（）友

32、情！予人玫瑰手留余香！結(jié)點(diǎn) 5 的為 2，結(jié)點(diǎn) 9 的為 1，和為 37Problem Id: 1817Submit time: 2010-04-14 10:49:17 User_id: bupt111310352 Memory:96KTime:16MS Language:G+Result:AcceptedCode*/#include <stdio.h>/ 結(jié)點(diǎn)的數(shù)據(jù)結(jié)構(gòu)typedef struct HNodeunsigned int weight; unsigned int parent;/ 父結(jié)點(diǎn)/在 m_Node 的 0 到 m_Cycle 之間尋找 parent并在數(shù)組最后

33、存放這兩個(gè)結(jié)點(diǎn)形成的父結(jié)點(diǎn)指針為 0且最小的兩個(gè)結(jié)點(diǎn)void Select( HNode* m_Node, int m_Cycle )/ 循環(huán)數(shù)int temp;/下最小的兩個(gè)數(shù)的位置int m_FirstMin_Pos = -1, m_SecondMin_Pos = -1;/ 最小的兩個(gè)數(shù)，初始賦值為題目所給最大值加 1 int m_FirstMin = 1001, m_SecondMin = 1001;for( temp = 0; temp <= m_Cycle ; + temp )if ( m_Node temp .parent = 0 )/如果結(jié)點(diǎn)則 FirstMinm_Firs

34、tMin 還小，將值傳遞給 m_Node_SecondMin, FirstMin 結(jié)點(diǎn)指向查出的最小值結(jié)點(diǎn)/該結(jié)點(diǎn)位置if( m_Node temp .weight < m_FirstMin )m_SecondMin = m_FirstMin;王道王道（）友情！予人玫瑰 手留余香！m_SecondMin_Pos = m_FirstMin_Pos; m_FirstMin = m_Node temp .weight;m_FirstMin_Pos = temp;else/ 如果 點(diǎn)FirstMin 大，但比 m_Node_SecondMin 小，/ 則 m_Node_SecondMin 指向

35、查出的最小值結(jié)點(diǎn)，并if( m_Node temp .weight < m_SecondMin )m_SecondMin = m_Node temp .weight; m_SecondMin_Pos = temp;位置/ 在 m_Cycle + 1 的位置為最小的兩個(gè)結(jié)點(diǎn)的和生成的結(jié)點(diǎn)m_Node + m_Cycle .weight = m_FirstMin + m_SecondMin;/ 兩個(gè)最小值結(jié)點(diǎn)的 parent 賦值m_Node m_FirstMin_Pos .parent = m_Cycle; m_Node m_SecondMin_Pos .parent = m_Cycle;int main()/ 葉結(jié)點(diǎn)的數(shù)目int m_Num_Count, m_Num_Count_Temp = 0; scanf( "%d", &m_Num_Count );/ 最后形成的樹總結(jié)點(diǎn)個(gè)數(shù)，結(jié)點(diǎn)個(gè)數(shù)* 2 - 1int m