枚举

先曝上sicily上的一道题吧，soj4190.

原题如下：

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4190. Prime Palindromes

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB

Description

The number 151 is a prime palindrome because it is both a prime number and a palindrome (it is the same number when read forward as backward). Write a program that finds all prime palindromes in the range of two supplied numbers a and b (5 <= a < b <= 100,000,000); both a and b are considered to be within the range .

Input

There are multiple test cases.

Each case contains two integers, a and b.

a=b=0 indicates the end of input.

Output

For each test case, output the list of palindromic primes in numerical order, one per line.

Sample Input

5 500
0 0

Sample Output

5
7
11
101
131
151
181
191
313
353
373
383

AC代码如下：

#include <iostream>
using namespace std;

const long long data[780]={3,5,7,11,101,131,151,181,191,313,353,373,383,727,757,787,797,919,929,10301,10501,10601,11311,11411,12421,12721,12821,13331,13831,13931,14341,14741,15451,15551,16061,16361,16561,16661,17471,17971,18181,18481,19391,19891,19991,30103,30203,30403,30703,30803,31013,31513,32323,32423,33533,34543,34843,35053,35153,35353,35753,36263,36563,37273,37573,38083,38183,38783,39293,70207,70507,70607,71317,71917,72227,72727,73037,73237,73637,74047,74747,75557,76367,76667,77377,77477,77977,78487,78787,78887,79397,79697,79997,90709,91019,93139,93239,93739,94049,94349,94649,94849,94949,95959,96269,96469,96769,97379,97579,97879,98389,98689,1003001,1008001,1022201,1028201,1035301,1043401,1055501,1062601,1065601,1074701,1082801,1085801,1092901,1093901,1114111,1117111,1120211,1123211,1126211,1129211,1134311,1145411,1150511,1153511,1160611,1163611,1175711,1177711,1178711,1180811,1183811,1186811,1190911,1193911,1196911,1201021,1208021,1212121,1215121,1218121,1221221,1235321,1242421,1243421,1245421,1250521,1253521,1257521,1262621,1268621,1273721,1276721,1278721,1280821,1281821,1286821,1287821,1300031,1303031,1311131,1317131,1327231,1328231,1333331,1335331,1338331,1343431,1360631,1362631,1363631,1371731,1374731,1390931,1407041,1409041,1411141,1412141,1422241,1437341,1444441,1447441,1452541,1456541,1461641,1463641,1464641,1469641,1486841,1489841,1490941,1496941,1508051,1513151,1520251,1532351,1535351,1542451,1548451,1550551,1551551,1556551,1557551,1565651,1572751,1579751,1580851,1583851,1589851,1594951,1597951,1598951,1600061,1609061,1611161,1616161,1628261,1630361,1633361,1640461,1643461,1646461,1654561,1657561,1658561,1660661,1670761,1684861,1685861,1688861,1695961,1703071,1707071,1712171,1714171,1730371,1734371,1737371,1748471,1755571,1761671,1764671,1777771,1793971,1802081,1805081,1820281,1823281,1824281,1826281,1829281,1831381,1832381,1842481,1851581,1853581,1856581,1865681,1876781,1878781,1879781,1880881,1881881,1883881,1884881,1895981,1903091,1908091,1909091,1917191,1924291,1930391,1936391,1941491,1951591,1952591,1957591,1958591,1963691,1968691,1969691,1970791,1976791,1981891,1982891,1984891,1987891,1988891,1993991,1995991,1998991,3001003,3002003,3007003,3016103,3026203,3064603,3065603,3072703,3073703,3075703,3083803,3089803,3091903,3095903,3103013,3106013,3127213,3135313,3140413,3155513,3158513,3160613,3166613,3181813,3187813,3193913,3196913,3198913,3211123,3212123,3218123,3222223,3223223,3228223,3233323,3236323,3241423,3245423,3252523,3256523,3258523,3260623,3267623,3272723,3283823,3285823,3286823,3288823,3291923,3293923,3304033,3305033,3307033,3310133,3315133,3319133,3321233,3329233,3331333,3337333,3343433,3353533,3362633,3364633,3365633,3368633,3380833,3391933,3392933,3400043,3411143,3417143,3424243,3425243,3427243,3439343,3441443,3443443,3444443,3447443,3449443,3452543,3460643,3466643,3470743,3479743,3485843,3487843,3503053,3515153,3517153,3528253,3541453,3553553,3558553,3563653,3569653,3586853,3589853,3590953,3591953,3594953,3601063,3607063,3618163,3621263,3627263,3635363,3643463,3646463,3670763,3673763,3680863,3689863,3698963,3708073,3709073,3716173,3717173,3721273,3722273,3728273,3732373,3743473,3746473,3762673,3763673,3765673,3768673,3769673,3773773,3774773,3781873,3784873,3792973,3793973,3799973,3804083,3806083,3812183,3814183,3826283,3829283,3836383,3842483,3853583,3858583,3863683,3864683,3867683,3869683,3871783,3878783,3893983,3899983,3913193,3916193,3918193,3924293,3927293,3931393,3938393,3942493,3946493,3948493,3964693,3970793,3983893,3991993,3994993,3997993,3998993,7014107,7035307,7036307,7041407,7046407,7057507,7065607,7069607,7073707,7079707,7082807,7084807,7087807,7093907,7096907,7100017,7114117,7115117,7118117,7129217,7134317,7136317,7141417,7145417,7155517,7156517,7158517,7159517,7177717,7190917,7194917,7215127,7226227,7246427,7249427,7250527,7256527,7257527,7261627,7267627,7276727,7278727,7291927,7300037,7302037,7310137,7314137,7324237,7327237,7347437,7352537,7354537,7362637,7365637,7381837,7388837,7392937,7401047,7403047,7409047,7415147,7434347,7436347,7439347,7452547,7461647,7466647,7472747,7475747,7485847,7486847,7489847,7493947,7507057,7508057,7518157,7519157,7521257,7527257,7540457,7562657,7564657,7576757,7586857,7592957,7594957,7600067,7611167,7619167,7622267,7630367,7632367,7644467,7654567,7662667,7665667,7666667,7668667,7669667,7674767,7681867,7690967,7693967,7696967,7715177,7718177,7722277,7729277,7733377,7742477,7747477,7750577,7758577,7764677,7772777,7774777,7778777,7782877,7783877,7791977,7794977,7807087,7819187,7820287,7821287,7831387,7832387,7838387,7843487,7850587,7856587,7865687,7867687,7868687,7873787,7884887,7891987,7897987,7913197,7916197,7930397,7933397,7935397,7938397,7941497,7943497,7949497,7957597,7958597,7960697,7977797,7984897,7985897,7987897,7996997,9002009,9015109,9024209,9037309,9042409,9043409,9045409,9046409,9049409,9067609,9073709,9076709,9078709,9091909,9095909,9103019,9109019,9110119,9127219,9128219,9136319,9149419,9169619,9173719,9174719,9179719,9185819,9196919,9199919,9200029,9209029,9212129,9217129,9222229,9223229,9230329,9231329,9255529,9269629,9271729,9277729,9280829,9286829,9289829,9318139,9320239,9324239,9329239,9332339,9338339,9351539,9357539,9375739,9384839,9397939,9400049,9414149,9419149,9433349,9439349,9440449,9446449,9451549,9470749,9477749,9492949,9493949,9495949,9504059,9514159,9526259,9529259,9547459,9556559,9558559,9561659,9577759,9583859,9585859,9586859,9601069,9602069,9604069,9610169,9620269,9624269,9626269,9632369,9634369,9645469,9650569,9657569,9670769,9686869,9700079,9709079,9711179,9714179,9724279,9727279,9732379,9733379,9743479,9749479,9752579,9754579,9758579,9762679,9770779,9776779,9779779,9781879,9782879,9787879,9788879,9795979,9801089,9807089,9809089,9817189,9818189,9820289,9822289,9836389,9837389,9845489,9852589,9871789,9888889,9889889,9896989,9902099,9907099,9908099,9916199,9918199,9919199,9921299,9923299,9926299,9927299,9931399,9932399,9935399,9938399,9957599,9965699,9978799,9980899,9981899,9989899};

int main(){
    int a,b;
    while(1){
        cin>>a>>b;
        if(a==0&&b==0){
            break;
        }
        for(int i=0;i<=780;i++){
            if(data[i]>=a&&data[i]<=b){
                cout<<data[i]<<endl;
            }
        }
    }    
} 

各位看官，看到这想必都有些懊恼，这不是作弊嘛，分明是打表出来的。。呵呵。。但是没办法，数据量实在太大，这也是无奈之举。

这里也贴出用C++文件操作，得到这些符合要求的数据的代码：

#include <iostream>
#include <cmath>
#include <fstream>
using namespace std;

bool ispalindrome(int a){
    int temp=0,temp1=a;    
    while(a>0){
        temp=temp*10+a%10;
        a=a/10;
    }
    return temp+a==temp1;
}


bool isprime(int a){
    int temp=floor(sqrt(a) +0.5);
    for(int i=2;i<=temp;i++){
        if(a%i==0) return 0;
    }
    return 1;
} 

int main(){
    ifstream in("a.txt");  //a.txt中存放的是由 ispalindrome &&（i%2!=0）筛选出来的数据，共11109个。 
    ofstream out("b.txt");
    long long data[11109];
    int count=0;
    for(int i=0;i<11109;i++){
        in>>data[i];
        if(isprime(data[i])){
            count++;
            out<<data[i]<<",";
        }
    }
    cout<<count;
    in.close();
    out.close();
}

此外，POJ1064也是枚举（二分法）的一个好例子，这里借花献佛，搬来别人的代码也总结一下：

/* *  poj 1064 Cable master

    题目大意:
        给定一系列长度的电缆，需要截取成等长的K份，求所能截取的最大长度值。

    解题思路:
        1、二分法枚举所有可能的长度，求得最大长度值 -- 所谓可能长度，即该长度能够产生多于或等于K份


    数学模型: 
        该题之所以可以采用二分法枚举，在于所能产生的电缆份数随长度的递增而单调非增，即有如下

        f(x)表示长度为x时产生的份数，f(x)随x的变化单调非增。 -- 所有单调模型均可采用二分枚举! 

    注意:
        本题WA的主要原因是精度问题，看discuss有人说输入数据+0.005即可满足AC条件，试了一下果然
        BT测试数据:

        4 2540
        8.02
        7.43
        4.57
        5.39

        =>0.01

        4 2542
        8.02
        7.43
        4.57
        5.39

        =>0.00
        
        另: 解决精度问题，也可以将输入数据乘以100换算成整数进行计算。

        2 2
        1.02
        2.33

        => 1.16 // 并没有说一定要把所有的电缆都用上，哎，这里WA了N次

*/

#include <iostream>
#include <cstdio>
#include <cstdlib>

namespace {
    using namespace std;

    const int N_MAX = 10000;
    int cables[N_MAX];

    int n;
    int cut_numbers(int len)
    {
        int count = 0;
        for (int i=0; i<n; i++)
        {
            count += cables[i]/len;
        }

        return count;
    }
}

#define ROUND_DOWN(x) 
    ((x)+0.005)

int main()
{
    int K;
    scanf("%d%d", &n, &K);

    double t;
    long long s=0, l=1, r=0; // l从1开始取
    for (int i=0; i<n; i++ )
    {
        scanf("%lf", &t);
        cables[i] = ROUND_DOWN(t)*100; // 4舍5入
        s += cables[i];

        if (cables[i]>r) r=cables[i]; // 这里千万注意，最大可能取值并不是读入的最小
    }

    if (s/K<r) r=s/K; // 如果平均值小于r，则r取平均

    // 二分查找
    int len=0;
    while (l<=r)
    {
        int q = (l+r)/2; // 从1取l，这里得到的q绝对不会是0
        if (cut_numbers(q) >= K)
        {
            l = q + 1;
            len = q;
        }
        else
        {
            r = q - 1;
        }
    }

    printf("%0.2lf
", len*0.01);
    
    return 0;
}