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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(kāi)發(fā)者計(jì)劃使來(lái)自于不同組織的開(kāi)發(fā)者,能夠在北電網(wǎng)絡(luò)的平臺(tái)上開(kāi)發(fā)圓滿(mǎn)的補(bǔ)充業(yè)務(wù)。北電網(wǎng)絡(luò)符合工業(yè)標(biāo)準(zhǔn)的開(kāi)放接口,為補(bǔ)充業(yè)務(wù)的開(kāi)展引入了無(wú)數(shù)商機(jī),開(kāi)發(fā)者計(jì)劃為不同層面的開(kāi)發(fā)者提供不同等級(jí)的資格,資格的劃分還考慮到以下因素:補(bǔ)充業(yè)務(wù)與北電網(wǎng)絡(luò)平臺(tái)的集合程度,開(kāi)發(fā)者團(tuán)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(gè)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(biāo)為(0,0) “7”的坐標(biāo)為(-1,-1) 編寫(xiě)一個(gè)小程序,使程序做到輸入坐標(biāo)(X,Y)之后顯示出相應(yīng)的數(shù)字。
3、#include
//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(gè)程序,當(dāng)已知一個(gè)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
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