int isPrime(int n) {              // (1)
    int i, sqrtn = sqrt(n + 1e-6);
    if(n <= 1) {
        return 0;                 // (2)
    }
    for(i = 2; i <= sqrtn; ++i) {
        if(n % i == 0) {          // (3)
            return 0;
        }
    }
    return 1;                     // (4)
}

其中n要加1e-6是因为，计算机内部存储数字是用的二进制，二进制无法精确表示十进制的小数会有误差，加上这个进行误差的消除。其实这个等式暗藏了浮点转整形的转换，你有看到么？

课后习题详解

回文素数

866. 回文素数https://leetcode-cn.com/problems/prime-palindrome/

?求出大于或等于?N?的最小回文素数。

思路1

利用一个函数来判断是否是回文。然后再判断是否是素数返回结果就好了。

顺序不能反，因为回文数判断快很多只要O(1)而素数需要1/2指数量级。

bool huiwen(int n){
    int s[10],i;
    for(i = 0;n;++i){
        s[i] = n%10;
        n = n / 10;
    }
    for(int j = 0;j < i / 2;++j)
        if(s[j] != s[i - 1 -j]) return false;
    return true;
}
int primePalindrome(int n){
    if(n == 1) return 2;
    int ans;
    for(int i = n; ; i++){
        int sqrti = sqrt(i + 1e-6);
        bool flag = true;
        //printf("%d",sqrti);
        for(int j = 2;j <= sqrti; ++j)
            if(i % j == 0){
                //printf("123");
                flag = false;
                break;
            }
        if(flag && huiwen(i)){
            ans = i;
            break;
        }
    }
    return ans;
}

结果分析1

超时！惊不惊喜，意不意外，然后去看了一下官方的题解，给出的解释是8位的数字没有素数。

我就emmmmm？我哪知道去？这也太扯了吧，于是优化思路形成第二种。

?思路2

我只要根据当前数字，找到大于他的最小回文数字，然后判断是否是素数，然后迭代就能找到满足题目的解了对不对。

所以有两个步骤

1.找对应的下一个回文数

其实比较简单，我就拿到前面的一半对称到后面就可以生成了，但是这样不能保证比我给的数字大，所以做个判断，如果不大我就把前面的数字加一就好了。但是我要有一个根据前面的一般数字生成后面数字的规则。👇k是代表是奇数位还是偶数位的

int huiwen(int n,int k){
    int s[10],ans = n,wei = 0;
    while(n){
        s[wei++] = n %10;
        n /= 10;
    }
    for(int i = k?1:0; i < wei; i++){
        ans *= 10;
        ans += s[i];
    }
    return ans;
}

然后就是返回回文数字的函数，要注意的是如果给的值是边界也就是几个9的话需要进行+1再寻找。

int nexthuiwen(int n){
    if(n % 10 == 9) n++;//边界条件
    int s[10],ans,wei = 0,temp =n;
    while(n){        //判断位数
        s[wei++] = n % 10;
        n /= 10;
    }
    if(wei % 2 == 1){
        int wei1 = (wei - 1) / 2;
        for(int i = temp /pow(10,wei1);;i++){//截取前一半
            ans = huiwen(i,1);
            if(ans > temp) break; 
        }
    }
    else{
        int wei2 = wei / 2;
        for(int i = temp/pow(10,wei2);;i++){//截取前一半
            ans = huiwen(i,0);
            if(ans > temp)  break;
        }
    }
    return ans;    //返回回文数字
}

完整代码

其实很简单的，就是根据数字找下一个回文数，然后判断回文数是不是素数，如果是就返回。

int huiwen(int n,int k){
    int s[10],ans = n,wei = 0;
    while(n){
        s[wei++] = n %10;
        n /= 10;
    }
    for(int i = k?1:0; i < wei; i++){
        ans *= 10;
        ans += s[i];
    }
    return ans;
}
int nexthuiwen(int n){
    if(n % 10 == 9) n++;
    int s[10],ans,wei = 0,temp =n;
    while(n){
        s[wei++] = n % 10;
        n /= 10;
    }
    if(wei % 2 == 1){
        int wei1 = (wei - 1) / 2;
        for(int i = temp /pow(10,wei1);;i++){
            ans = huiwen(i,1);
            if(ans > temp) break; 
        }
    }
    else{
        int wei2 = wei / 2;
        for(int i = temp/pow(10,wei2);;i++){
            ans = huiwen(i,0);
            if(ans > temp)  break;
        }
    }
    return ans;
}
int primePalindrome(int n){
    if(n == 1) return 2;
    int ans = 0;
    for(int i = n - 1; ; ){
        i = nexthuiwen(i); 
        //素数判定
        int sqrti = sqrt(i + 1e-6);
        bool flag = true;
        for(int j = 2;j <= sqrti; ++j)//
            if(i % j == 0){
                //printf("123");
                flag = false;
                break;
            }
        if(flag){
            ans = i;
            break;
        }
    
    }
    return ans;
}

?结果分析

完美！

?丑数?剑指 Offer 49. 丑数https://leetcode-cn.com/problems/chou-shu-lcof/

?思路1

因为每个丑数都是通过一些2一些3一些5生成的，那么，如果我想要生成下一个丑数肯定是是在之前的所有丑数分别乘2乘3乘5得到的比上一个丑数大的数中最小的。(多绕绕肯定能绕过来)

对应算法就是我有三个记录位置的指针，这三个指针分别指向x2、x3、x5后刚好比我之前生成的丑数大的元素，然后下一个元素就会在这三个数中产生。

    for(int i = 1; i < n; i++){
        int temp = min(*p2 * 2,*p3 * 3,*p5 * 5);
        num[i] = temp;
        while(*p2 * 2 <= temp)   p2++;
        while(*p3 * 3 <= temp)   p3++;
        while(*p5 * 5 <= temp)   p5++;
    }

完整代码1

c好像没有取三个元素最小的，就随手写了个。不复杂。

int min(int a,int b, int c){
    if(a > b)
        if(b > c)   return c;
        else return b;
    else
        if(b < c)   return a;
        else if(a > c)  return c;
        else return a;
}
int nthUglyNumber(int n){
    int num[1690] = {1};
    int *p2 = num, *p3= num, *p5 = num;
    for(int i = 1; i < n; i++){
        int temp = min(*p2 * 2,*p3 * 3,*p5 * 5);
        printf("%d\n",temp);
        printf("%d %d %d\n",*p2 * 2,*p3 * 3,*p5 * 5);
        num[i] = temp;
        while(*p2 * 2 <= temp)   p2++;
        while(*p3 * 3 <= temp)   p3++;
        while(*p5 * 5 <= temp)   p5++;
        printf("%d %d %d\n",*p2,*p3,*p5);
    }
    return num[n - 1];
}

结果分析1

这种情况这么低的排名。。我怀疑他们作弊，所以

我们也作弊把！?

耍赖方法

思路

其实很简答，就1960个数字，保存到数组里查多快啊。

你如果问你怎么知道这1960个数字。让计算机算啊。你就是写出来指数的函数用你电脑算出来，你都能赢0.0

#include<cstdio>

int min(int a,int b, int c){
    if(a > b)
        if(b > c)   return c;
        else return b;
    else
        if(b < c)   return a;
        else if(a > c)  return c;
        else return a;
}
int nthUglyNumber(int n){
    int num[1690] = {1};
    int *p2 = num, *p3= num, *p5 = num;
    for(int i = 1; i < n; i++){
        int temp = min(*p2 * 2,*p3 * 3,*p5 * 5);
        printf("%d,",temp);
        //printf("%d %d %d\n",*p2 * 2,*p3 * 3,*p5 * 5);
        num[i] = temp;
        while(*p2 * 2 <= temp)   p2++;
        while(*p3 * 3 <= temp)   p3++;
        while(*p5 * 5 <= temp)   p5++;
        //printf("%d %d %d\n",*p2,*p3,*p5);
    }
    return num[n - 1];
}
int main(){
    freopen("output.txt","w",stdout);
    nthUglyNumber(1690);
    return 0;
}

int nthUglyNumber(int n){
    int num[] ={0,1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36,40,45,48,50,54,60,64,72,75,80,81,90,96,100,108,120,125,128,135,144,150,160,162,180,192,200,216,225,240,243,250,256,270,288,300,320,324,360,375,384,400,405,432,450,480,486,500,512,540,576,600,625,640,648,675,720,729,750,768,800,810,864,900,960,972,1000,1024,1080,1125,1152,1200,1215,1250,1280,1296,1350,1440,1458,1500,1536,1600,1620,1728,1800,1875,1920,1944,2000,2025,2048,2160,2187,2250,2304,2400,2430,2500,2560,2592,2700,2880,2916,3000,3072,3125,3200,3240,3375,3456,3600,3645,3750,3840,3888,4000,4050,4096,4320,4374,4500,4608,4800,4860,5000,5120,5184,5400,5625,5760,5832,6000,6075,6144,6250,6400,6480,6561,6750,6912,7200,7290,7500,7680,7776,8000,8100,8192,8640,8748,9000,9216,9375,9600,9720,10000,10125,10240,10368,10800,10935,11250,11520,11664,12000,12150,12288,12500,12800,12960,13122,13500,13824,14400,14580,15000,15360,15552,15625,16000,16200,16384,16875,17280,17496,18000,18225,18432,18750,19200,19440,19683,20000,20250,20480,20736,21600,21870,22500,23040,23328,24000,24300,24576,25000,25600,25920,26244,27000,27648,28125,28800,29160,30000,30375,30720,31104,31250,32000,32400,32768,32805,33750,34560,34992,36000,36450,36864,37500,38400,38880,39366,40000,40500,40960,41472,43200,43740,45000,46080,46656,46875,48000,48600,49152,50000,50625,51200,51840,52488,54000,54675,55296,56250,57600,58320,59049,60000,60750,61440,62208,62500,64000,64800,65536,65610,67500,69120,69984,72000,72900,73728,75000,76800,77760,78125,78732,80000,81000,81920,82944,84375,86400,87480,90000,91125,92160,93312,93750,96000,97200,98304,98415,100000,101250,102400,103680,104976,108000,109350,110592,112500,115200,116640,118098,120000,121500,122880,124416,125000,128000,129600,131072,131220,135000,138240,139968,140625,144000,145800,147456,150000,151875,153600,155520,156250,157464,160000,162000,163840,164025,165888,168750,172800,174960,177147,180000,182250,184320,186624,187500,192000,194400,196608,196830,200000,202500,204800,207360,209952,216000,218700,221184,225000,230400,233280,234375,236196,240000,243000,245760,248832,250000,253125,256000,259200,262144,262440,270000,273375,276480,279936,281250,288000,291600,294912,295245,300000,303750,307200,311040,312500,314928,320000,324000,327680,328050,331776,337500,345600,349920,354294,360000,364500,368640,373248,375000,384000,388800,390625,393216,393660,400000,405000,409600,414720,419904,421875,432000,437400,442368,450000,455625,460800,466560,468750,472392,480000,486000,491520,492075,497664,500000,506250,512000,518400,524288,524880,531441,540000,546750,552960,559872,562500,576000,583200,589824,590490,600000,607500,614400,622080,625000,629856,640000,648000,655360,656100,663552,675000,691200,699840,703125,708588,720000,729000,737280,746496,750000,759375,768000,777600,781250,786432,787320,800000,810000,819200,820125,829440,839808,843750,864000,874800,884736,885735,900000,911250,921600,933120,937500,944784,960000,972000,983040,984150,995328,1000000,1012500,1024000,1036800,1048576,1049760,1062882,1080000,1093500,1105920,1119744,1125000,1152000,1166400,1171875,1179648,1180980,1200000,1215000,1228800,1244160,1250000,1259712,1265625,1280000,1296000,1310720,1312200,1327104,1350000,1366875,1382400,1399680,1406250,1417176,1440000,1458000,1474560,1476225,1492992,1500000,1518750,1536000,1555200,1562500,1572864,1574640,1594323,1600000,1620000,1638400,1640250,1658880,1679616,1687500,1728000,1749600,1769472,1771470,1800000,1822500,1843200,1866240,1875000,1889568,1920000,1944000,1953125,1966080,1968300,1990656,2000000,2025000,2048000,2073600,2097152,2099520,2109375,2125764,2160000,2187000,2211840,2239488,2250000,2278125,2304000,2332800,2343750,2359296,2361960,2400000,2430000,2457600,2460375,2488320,2500000,2519424,2531250,2560000,2592000,2621440,2624400,2654208,2657205,2700000,2733750,2764800,2799360,2812500,2834352,2880000,2916000,2949120,2952450,2985984,3000000,3037500,3072000,3110400,3125000,3145728,3149280,3188646,3200000,3240000,3276800,3280500,3317760,3359232,3375000,3456000,3499200,3515625,3538944,3542940,3600000,3645000,3686400,3732480,3750000,3779136,3796875,3840000,3888000,3906250,3932160,3936600,3981312,4000000,4050000,4096000,4100625,4147200,4194304,4199040,4218750,4251528,4320000,4374000,4423680,4428675,4478976,4500000,4556250,4608000,4665600,4687500,4718592,4723920,4782969,4800000,4860000,4915200,4920750,4976640,5000000,5038848,5062500,5120000,5184000,5242880,5248800,5308416,5314410,5400000,5467500,5529600,5598720,5625000,5668704,5760000,5832000,5859375,5898240,5904900,5971968,6000000,6075000,6144000,6220800,6250000,6291456,6298560,6328125,6377292,6400000,6480000,6553600,6561000,6635520,6718464,6750000,6834375,6912000,6998400,7031250,7077888,7085880,7200000,7290000,7372800,7381125,7464960,7500000,7558272,7593750,7680000,7776000,7812500,7864320,7873200,7962624,7971615,8000000,8100000,8192000,8201250,8294400,8388608,8398080,8437500,8503056,8640000,8748000,8847360,8857350,8957952,9000000,9112500,9216000,9331200,9375000,9437184,9447840,9565938,9600000,9720000,9765625,98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    return num[n];
}

下面的数组就是上面的函数生成的。直接查表就好了。