m*d
2 楼
最近onsite的时候刚拿到一道面试题,what is a computational fast way to divide
an integer by 7.
有大神有思路么?
求个解答算法或者代码?
an integer by 7.
有大神有思路么?
求个解答算法或者代码?
L*x
3 楼
rt
xiexie!
xiexie!
k*9
4 楼
亲戚来访 想给买个作礼物
有没有这个手机的用户能提供一下经验 比如能否用4G 谢谢
有没有这个手机的用户能提供一下经验 比如能否用4G 谢谢
n*s
10 楼
a/7=a/8+(a/7)/8
h*b
11 楼
没有国内lte频段,最多用用联通3g,直接在国内买7c就行了,7x和7c一个祖宗
l*s
13 楼
private static int divideBy7(int num){
if (num == int.MinValue) return -divideBy7(int.MaxValue);
if (num < 0) return -divideBy7(-num);
int result = 0;
while (num >= 7){
int subResult = 1, div = 7;
while (((div << 3) | 7) > 0 && num >= ((div << 3) | 7)){
div = (div << 3) | 7;
subResult = subResult << 3 | 1;
}
while ((div << 1) > 0 && num >= (div << 1)){
div <<= 1; subResult <<= 1;
}
num -= div; result += subResult;
}
return result;
}
if (num == int.MinValue) return -divideBy7(int.MaxValue);
if (num < 0) return -divideBy7(-num);
int result = 0;
while (num >= 7){
int subResult = 1, div = 7;
while (((div << 3) | 7) > 0 && num >= ((div << 3) | 7)){
div = (div << 3) | 7;
subResult = subResult << 3 | 1;
}
while ((div << 1) > 0 && num >= (div << 1)){
div <<= 1; subResult <<= 1;
}
num -= div; result += subResult;
}
return result;
}
k*9
14 楼
谢谢楼上2位
那就算了
有没有差不多价位的但是支持中美频段的?(中国移动,美国att)
那就算了
有没有差不多价位的但是支持中美频段的?(中国移动,美国att)
g*7
15 楼
int div7(int num)
{
int Q, R, sum;
// got this from post #4
if ( num < 0 )
return (-div7(-num));
R = num;
sum = 0;
while (R & 0xFFF8 )
{
Q = R >> 3;
sum += Q;
R = R & 0x7;
R = R + Q;
}
if ( R == 7 )
sum++;
return sum;
}
{
int Q, R, sum;
// got this from post #4
if ( num < 0 )
return (-div7(-num));
R = num;
sum = 0;
while (R & 0xFFF8 )
{
Q = R >> 3;
sum += Q;
R = R & 0x7;
R = R + Q;
}
if ( R == 7 )
sum++;
return sum;
}
g*7
17 楼
// one fewer calculation
int div7(int num)
{
int Q, R, sum;
// got this from post #4
if ( num < 0 )
return (-div7(-num));
R = num;
sum = 0;
Q = R >> 3;
while (Q)
{
sum += Q;
R = R & 0x7;
R = R + Q;
Q = R >> 3;
}
if ( R == 7 )
sum++;
return sum;
}
int div7(int num)
{
int Q, R, sum;
// got this from post #4
if ( num < 0 )
return (-div7(-num));
R = num;
sum = 0;
Q = R >> 3;
while (Q)
{
sum += Q;
R = R & 0x7;
R = R + Q;
Q = R >> 3;
}
if ( R == 7 )
sum++;
return sum;
}
s*g
19 楼
按位置1后乘7(右移三位再减自己),看是不是比被除数大,大的话该位置回零
int div7(int num) {
if (num < 0)
return -div7(-num);
int H = num >> 2;
int bits = 0;
while(H > 0){
H >>= 1;
bits ++;
}
int res = 0;
for( ; bits >= 0 ; bits--) {
int mask = 1 << bits;
res = res | mask;
int a = (res << 3) - res;
if (a == num)
return res;
else if (a > num)
res = res ^ mask;
}
return res;
}
int div7(int num) {
if (num < 0)
return -div7(-num);
int H = num >> 2;
int bits = 0;
while(H > 0){
H >>= 1;
bits ++;
}
int res = 0;
for( ; bits >= 0 ; bits--) {
int mask = 1 << bits;
res = res | mask;
int a = (res << 3) - res;
if (a == num)
return res;
else if (a > num)
res = res ^ mask;
}
return res;
}
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