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名企面試試題——廣東北電
英文筆試題1. Tranlation (Mandatory)
CDMA venders have worked hard to give CDMA roaming capabilities via the development of RUIM-essentially, a SIM card for CDMA handsets currently being deployed in China for new CDMA operator China Unicom. Korean cellco KTF demonstrated earlier this year the ability to roam between GSM and CDMA using such cards.However,only the card containing the user’s service data can roam-not the CDMA handset or the user’s number (except via call forwarding).
2. Programming (Mandatory)
Linked list
a. Implement a linked list for integers,which supports the insertafter (insert a node after a specified node) and removeafter (remove the node after a specified node) methods;
b. Implement a method to sort the linked list to descending order.
3. Debugging (Mandatory)
a. For each of the following recursive methods,enter Y in the answer box if themethod terminaters (assume i=5), Otherwise enter N.
static int f(int i){
return f(i-1)*f(i-1);
}
Ansewr:
static int f(int i){
if(i==0){return 1;}
else {return f(i-1)*f(i-1);}
}
Ansewr:
static int f(int i){
if(i==0){return 1;}
else {return f(i-1)*f(i-2);}
}
Ansewr:
b. There are two errors in the following JAVA program:
static void g(int i){
if(i==1){return;}
if(i%2==0){g(i/2);return;}
else {g(3*i);return;}
}
please correct them to make sure we can get the printed-out result as below:
3 10 5 16 8 4 2 1
中文筆試題
1．漢譯英
北電網(wǎng)絡(luò)的開發(fā)者計劃使來自于不同組織的開發(fā)者，能夠在北電網(wǎng)絡(luò)的平臺上開發(fā)圓滿的補充業(yè)務(wù)。北電網(wǎng)絡(luò)符合工業(yè)標準的開放接口，為補充業(yè)務(wù)的開展引入了無數(shù)商機，開發(fā)者計劃為不同層面的開發(fā)者提供不同等級的資格，資格的劃分還考慮到以下因素：補充業(yè)務(wù)與北電網(wǎng)絡(luò)平臺的集合程度，開發(fā)者團體與北電網(wǎng)絡(luò)的合作關(guān)系，等等。
2．編程
將整數(shù)轉(zhuǎn)換成字符串：void itoa(int,char);
例如itoa(-123,s[])則s=“-123”;
網(wǎng)易
1、10個人分成4組 有幾種分法？
2、如圖：
7 8 9 10
6 1 2 11
5 4 3 12
16 15 14 13
設(shè)“1”的坐標為（0，0） “7”的坐標為（－1，－1） 編寫一個小程序，使程序做到輸入坐標（X,Y）之后顯示出相應(yīng)的數(shù)字。
3、#include<stdio.h>
//example input and output
//in 1 2 3 out 1 3 1
//in 123456789 2 100 out 123456789 100 21
long mex(long a,long b,long c)
{ long d;
if(b==0) return 0;
if(b==1) return a%c;
d=mex(a,b/2,c)； d*=d;這里忘了;d*=mex(a,b%2,c);d%=c;
return d;
}
int main(void)
{ long x,y,z;
while(1)
{ if(scanf(%d %d %d,&x,&y,&z)>3) return 0;
if(x<0) { printf("too small ");continue;}
if(y<0) { printf("too small ");continue;}
if(z<1) { printf("too small ");continue;}
if(y>z) { printf("too big ");continue;}
if(z>1000000010) {printf("too big ");continue}
printf(%d %d %d,x,z,mex(x,y,z);
}}
根據(jù)這個程序，當已知一個輸入，算出輸出，如：輸入 1 3 1 則輸出 1 2 3 輸入 123456789 100 21 輸出 123456789 2 100 http://www.ipr-jzsc.com/
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