你应该从一开始就去掉N的2次方部分。

第二次编辑下面m的神奇表达式应该是

m = N - (N & (N-1));

而不是书面的

第二次编辑结束

m = N & (N-1); // the lawest bit of N
N /= m;
byte = N & 0x0F;
if ((m % 2) || (byte !=1 && byte !=9))
  return false;

第一次编辑：

轻微改进：

m = N & (N-1); // the lawest bit of N
N /= m;
if ((m % 2) || (N & 0x07 != 1))
  return false;

第一次编辑结束

现在像往常一样继续。这样，当你到达浮点部分时，你已经去掉了所有2次方部分为奇数（大约一半）的数字，然后你只考虑剩下的1/8。也就是说，你在6%的数字上运行浮点部分。

可能是该问题的最佳算法是快速整数平方根算法https://stackoverflow.com/a/51585204/5191852

@Kde声称牛顿法的三次迭代对于32位整数的精度为±1就足够了。当然，64位整数需要更多的迭代，可能是6或7。

如果最后的X位数字是N，那么应该可以更有效地包装“不能是完美的正方形”！我将使用java 32位int，并生成足够的数据来检查数字的最后16位，即2048个十六进制int值。

...

好吧。要么我遇到了一些超出我理解范围的数论，要么我的代码中有一个错误。无论如何，以下是代码：

public static void main(String[] args) {
    final int BITS = 16;

    BitSet foo = new BitSet();

    for(int i = 0; i< (1<<BITS); i++) {
        int sq = (i*i);
        sq = sq & ((1<<BITS)-1);
        foo.set(sq);
    }

    System.out.println("int[] mayBeASquare = {");

    for(int i = 0; i< 1<<(BITS-5); i++) {
        int kk = 0;
        for(int j = 0; j<32; j++) {
            if(foo.get((i << 5) | j)) {
                kk |= 1<<j;
            }
        }
        System.out.print("0x" + Integer.toHexString(kk) + ", ");
        if(i%8 == 7) System.out.println();
    }
    System.out.println("};");
}

结果如下：

（ed：由于pretify.js性能不佳而取消；查看修订历史以查看。）

我找到了一种比你的6位+卡马克+sqrt代码快35%的方法，至少在我的CPU（x86）和编程语言（C/C++）中是这样。您的结果可能会有所不同，特别是因为我不知道Java因素将如何发挥作用。

我的方法有三个方面：

首先，过滤掉显而易见的答案。这包括负数和查看最后4位。（我发现看最后六个没有帮助。）我也回答0是。（在阅读下面的代码时，请注意我的输入是int64x。）如果（x<0||（x&2）||（（x&7）==5）||（x&11）==8））return false；如果（x==0）返回true；接下来，检查它是否是模255＝3*5*17的平方。因为这是三个不同素数的乘积，所以只有大约1/8的模255残数是正方形。然而，根据我的经验，调用模运算符（%）的成本比获得的收益更高，因此我使用涉及255＝2^8-1的位技巧来计算残差。（不管是好是坏，我没有使用从单词中读取单个字节的技巧，只使用按位和和移位。）int64 y=x；y=（y&4294967295LL）+（y>>32）；y=（y&65535）+（y>>16）；y=（y&255）+（（y>>8）&255）=（y>>16）；//此时，y介于0和511之间。更多的代码可以进一步减少它。为了实际检查残差是否为正方形，我在预先计算的表中查找答案。如果（bad255[y]）return false；//然而，我只使用大小为512的表最后，尝试使用类似于Hensel引理的方法计算平方根。（我不认为它直接适用，但经过一些修改后可以使用。）在此之前，我用二进制搜索将2的所有幂除以：如果（（x&4294967295LL）==0）x>>=32；如果（（x&65535）==0）x>>=16；如果（（x&255）==0）x>>=8；如果（（x&15）==0）x>>=4；如果（（x&3）==0）x>>=2；在这一点上，我们的数字是一个正方形，它必须是1模8。如果（（x&7）！=1)return false；亨塞尔引理的基本结构如下。（注意：未经测试的代码；如果不起作用，请尝试t=2或8。）int64 t=4，r=1；t<<=1；r+=（（x-r*r）&t）>>1；t<<=1；r+=（（x-r*r）&t）>>1；t<<=1；r+=（（x-r*r）&t）>>1；//重复此操作，直到t为2^33左右。如果需要，请使用循环。其思想是，在每次迭代时，将一位加到r上，即x的“当前”平方根；每个平方根都是精确的模2的一个越来越大的幂，即t/2。最后，r和t/2-r将是x模t/2的平方根。（注意，如果r是x的平方根，那么-r也是如此。即使是模数，这也是正确的，但要注意，对某些数进行模运算，事物可能会有2个以上的平方根；值得注意的是，这包括2的幂。）因为我们的实际平方根小于2^32，所以在这一点上，我们实际上可以检查r或t/2-r是否是真正的平方根。在我的实际代码中，我使用了以下修改的循环：整数64 r，t，z；r=开始[（x>>3）&1023]；做{z=x-r*r；如果（z==0）返回true；如果（z<0）return false；t=z&（-z）；r+=（z&t）>>1；如果（r>（t>>1））r=t-r；}而（t<=（1LL<<33））；这里的加速是通过三种方式获得的：预先计算的开始值（相当于循环的约10次迭代）、提前退出循环以及跳过一些t值。对于最后一部分，我看z=r-x*x，用一个小技巧将t设为2除以z的最大幂。这允许我跳过t值，这些值无论如何都不会影响r的值。在我的例子中，预先计算的起始值选取模8192的“最小正”平方根。

Even if this code doesn't work faster for you, I hope you enjoy some of the ideas it contains. Complete, tested code follows, including the precomputed tables.

typedef signed long long int int64;

int start[1024] =
{1,3,1769,5,1937,1741,7,1451,479,157,9,91,945,659,1817,11,
1983,707,1321,1211,1071,13,1479,405,415,1501,1609,741,15,339,1703,203,
129,1411,873,1669,17,1715,1145,1835,351,1251,887,1573,975,19,1127,395,
1855,1981,425,453,1105,653,327,21,287,93,713,1691,1935,301,551,587,
257,1277,23,763,1903,1075,1799,1877,223,1437,1783,859,1201,621,25,779,
1727,573,471,1979,815,1293,825,363,159,1315,183,27,241,941,601,971,
385,131,919,901,273,435,647,1493,95,29,1417,805,719,1261,1177,1163,
1599,835,1367,315,1361,1933,1977,747,31,1373,1079,1637,1679,1581,1753,1355,
513,1539,1815,1531,1647,205,505,1109,33,1379,521,1627,1457,1901,1767,1547,
1471,1853,1833,1349,559,1523,967,1131,97,35,1975,795,497,1875,1191,1739,
641,1149,1385,133,529,845,1657,725,161,1309,375,37,463,1555,615,1931,
1343,445,937,1083,1617,883,185,1515,225,1443,1225,869,1423,1235,39,1973,
769,259,489,1797,1391,1485,1287,341,289,99,1271,1701,1713,915,537,1781,
1215,963,41,581,303,243,1337,1899,353,1245,329,1563,753,595,1113,1589,
897,1667,407,635,785,1971,135,43,417,1507,1929,731,207,275,1689,1397,
1087,1725,855,1851,1873,397,1607,1813,481,163,567,101,1167,45,1831,1205,
1025,1021,1303,1029,1135,1331,1017,427,545,1181,1033,933,1969,365,1255,1013,
959,317,1751,187,47,1037,455,1429,609,1571,1463,1765,1009,685,679,821,
1153,387,1897,1403,1041,691,1927,811,673,227,137,1499,49,1005,103,629,
831,1091,1449,1477,1967,1677,697,1045,737,1117,1737,667,911,1325,473,437,
1281,1795,1001,261,879,51,775,1195,801,1635,759,165,1871,1645,1049,245,
703,1597,553,955,209,1779,1849,661,865,291,841,997,1265,1965,1625,53,
1409,893,105,1925,1297,589,377,1579,929,1053,1655,1829,305,1811,1895,139,
575,189,343,709,1711,1139,1095,277,993,1699,55,1435,655,1491,1319,331,
1537,515,791,507,623,1229,1529,1963,1057,355,1545,603,1615,1171,743,523,
447,1219,1239,1723,465,499,57,107,1121,989,951,229,1521,851,167,715,
1665,1923,1687,1157,1553,1869,1415,1749,1185,1763,649,1061,561,531,409,907,
319,1469,1961,59,1455,141,1209,491,1249,419,1847,1893,399,211,985,1099,
1793,765,1513,1275,367,1587,263,1365,1313,925,247,1371,1359,109,1561,1291,
191,61,1065,1605,721,781,1735,875,1377,1827,1353,539,1777,429,1959,1483,
1921,643,617,389,1809,947,889,981,1441,483,1143,293,817,749,1383,1675,
63,1347,169,827,1199,1421,583,1259,1505,861,457,1125,143,1069,807,1867,
2047,2045,279,2043,111,307,2041,597,1569,1891,2039,1957,1103,1389,231,2037,
65,1341,727,837,977,2035,569,1643,1633,547,439,1307,2033,1709,345,1845,
1919,637,1175,379,2031,333,903,213,1697,797,1161,475,1073,2029,921,1653,
193,67,1623,1595,943,1395,1721,2027,1761,1955,1335,357,113,1747,1497,1461,
1791,771,2025,1285,145,973,249,171,1825,611,265,1189,847,1427,2023,1269,
321,1475,1577,69,1233,755,1223,1685,1889,733,1865,2021,1807,1107,1447,1077,
1663,1917,1129,1147,1775,1613,1401,555,1953,2019,631,1243,1329,787,871,885,
449,1213,681,1733,687,115,71,1301,2017,675,969,411,369,467,295,693,
1535,509,233,517,401,1843,1543,939,2015,669,1527,421,591,147,281,501,
577,195,215,699,1489,525,1081,917,1951,2013,73,1253,1551,173,857,309,
1407,899,663,1915,1519,1203,391,1323,1887,739,1673,2011,1585,493,1433,117,
705,1603,1111,965,431,1165,1863,533,1823,605,823,1179,625,813,2009,75,
1279,1789,1559,251,657,563,761,1707,1759,1949,777,347,335,1133,1511,267,
833,1085,2007,1467,1745,1805,711,149,1695,803,1719,485,1295,1453,935,459,
1151,381,1641,1413,1263,77,1913,2005,1631,541,119,1317,1841,1773,359,651,
961,323,1193,197,175,1651,441,235,1567,1885,1481,1947,881,2003,217,843,
1023,1027,745,1019,913,717,1031,1621,1503,867,1015,1115,79,1683,793,1035,
1089,1731,297,1861,2001,1011,1593,619,1439,477,585,283,1039,1363,1369,1227,
895,1661,151,645,1007,1357,121,1237,1375,1821,1911,549,1999,1043,1945,1419,
1217,957,599,571,81,371,1351,1003,1311,931,311,1381,1137,723,1575,1611,
767,253,1047,1787,1169,1997,1273,853,1247,413,1289,1883,177,403,999,1803,
1345,451,1495,1093,1839,269,199,1387,1183,1757,1207,1051,783,83,423,1995,
639,1155,1943,123,751,1459,1671,469,1119,995,393,219,1743,237,153,1909,
1473,1859,1705,1339,337,909,953,1771,1055,349,1993,613,1393,557,729,1717,
511,1533,1257,1541,1425,819,519,85,991,1693,503,1445,433,877,1305,1525,
1601,829,809,325,1583,1549,1991,1941,927,1059,1097,1819,527,1197,1881,1333,
383,125,361,891,495,179,633,299,863,285,1399,987,1487,1517,1639,1141,
1729,579,87,1989,593,1907,839,1557,799,1629,201,155,1649,1837,1063,949,
255,1283,535,773,1681,461,1785,683,735,1123,1801,677,689,1939,487,757,
1857,1987,983,443,1327,1267,313,1173,671,221,695,1509,271,1619,89,565,
127,1405,1431,1659,239,1101,1159,1067,607,1565,905,1755,1231,1299,665,373,
1985,701,1879,1221,849,627,1465,789,543,1187,1591,923,1905,979,1241,181};

bool bad255[512] =
{0,0,1,1,0,1,1,1,1,0,1,1,1,1,1,0,0,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,
 1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,1,1,
 0,1,0,1,1,0,0,1,1,1,1,1,0,1,1,1,1,0,1,1,0,0,1,1,1,1,1,1,1,1,0,1,
 1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,0,1,1,1,0,1,1,1,1,0,0,1,1,1,1,1,1,
 1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,0,1,1,0,1,1,1,1,1,
 1,1,1,1,1,1,0,1,1,0,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1,
 1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,
 1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,
 0,0,1,1,0,1,1,1,1,0,1,1,1,1,1,0,0,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,
 1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,1,1,
 0,1,0,1,1,0,0,1,1,1,1,1,0,1,1,1,1,0,1,1,0,0,1,1,1,1,1,1,1,1,0,1,
 1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,0,1,1,1,0,1,1,1,1,0,0,1,1,1,1,1,1,
 1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,0,1,1,0,1,1,1,1,1,
 1,1,1,1,1,1,0,1,1,0,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1,
 1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,
 1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,
 0,0};

inline bool square( int64 x ) {
    // Quickfail
    if( x < 0 || (x&2) || ((x & 7) == 5) || ((x & 11) == 8) )
        return false;
    if( x == 0 )
        return true;

    // Check mod 255 = 3 * 5 * 17, for fun
    int64 y = x;
    y = (y & 4294967295LL) + (y >> 32);
    y = (y & 65535) + (y >> 16);
    y = (y & 255) + ((y >> 8) & 255) + (y >> 16);
    if( bad255[y] )
        return false;

    // Divide out powers of 4 using binary search
    if((x & 4294967295LL) == 0)
        x >>= 32;
    if((x & 65535) == 0)
        x >>= 16;
    if((x & 255) == 0)
        x >>= 8;
    if((x & 15) == 0)
        x >>= 4;
    if((x & 3) == 0)
        x >>= 2;

    if((x & 7) != 1)
        return false;

    // Compute sqrt using something like Hensel's lemma
    int64 r, t, z;
    r = start[(x >> 3) & 1023];
    do {
        z = x - r * r;
        if( z == 0 )
            return true;
        if( z < 0 )
            return false;
        t = z & (-z);
        r += (z & t) >> 1;
        if( r > (t  >> 1) )
            r = t - r;
    } while( t <= (1LL << 33) );
    
    return false;
}

static boolean isPerfectSquare (int input) {
  return Math.sqrt(input) == (int) Math.sqrt(input);
}

如果输入的平方根的整数值等于双倍值，则返回该值。这意味着它是一个整数，它将返回true。否则，将返回false。

考虑到一般的比特长度（尽管我在这里使用了特定的类型），我试图设计如下的简单算法。最初需要对0,1,2或<0进行简单而明显的检查。以下是简单的，因为它不试图使用任何现有的数学函数。大多数运算符可以用逐位运算符替换。我还没有用任何基准数据进行测试。我既不是数学专家，也不是计算机算法设计专家，我很乐意看到你们指出这个问题。我知道那里有很多改进的机会。

int main()
{
    unsigned int c1=0 ,c2 = 0;  
    unsigned int x = 0;  
    unsigned int p = 0;  
    int k1 = 0;  
    scanf("%d",&p);  
    if(p % 2 == 0) {  
        x = p/2; 
    }  
    else {  
        x = (p/2) +1;  
    }  
    while(x) 
    {
        if((x*x) > p) {  
            c1 = x;  
            x = x/2; 
        }else {  
            c2 = x;  
            break;  
        }  
    }  
    if((p%2) != 0)  
        c2++;

    while(c2 < c1) 
    {  
        if((c2 * c2 ) == p) {  
            k1 = 1;  
            break;  
        }  
        c2++; 
    }  
    if(k1)  
        printf("\n Perfect square for %d", c2);  
    else  
        printf("\n Not perfect but nearest to :%d :", c2);  
    return 0;  
}