时间跨度：01-07 15:52:16 - 01-09 18:19:02
(8号摸鱼去了)
强化知识点：
埃氏筛，分解质因数，快速幂，逆元
T1 快速幂||取余运算
题目链接：洛谷P1226
题目解析
快速幂模板题，按题意最后取余即可
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll fastpow(ll x,ll y,ll mod)
{
x %= mod;
ll res = 1;
while(y)
{
if(y&1)
res = res * x % mod;
y >>= 1;
x = x * x % mod;
}
return res;
}
int main()
{
ll b,p,k;
scanf("%lld %lld %lld",&b,&p,&k);
printf("%lld^%lld mod %lld=%lld",b,p,k,fastpow(b,p,k)%k);
return 0;
}
T2 线性筛素数
题目链接：洛谷P3383
题目解析
埃氏筛优化，保证每个合数只被筛一遍。将每个素数依次存入数组中，第k小的素数即为数组中第k-1个元素。
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e8+10;
int prime[MAXN];
ll num[MAXN];
void isPrime(ll n)
{
ll cnt = 0;
prime[0] = prime[1] = 1;
for(ll i = 2; i <= n; i++)
{
if(!prime[i])
num[cnt++] = i;
for(ll j = 0; j < cnt && i*num[j] < MAXN; j++)
{
prime[i*num[j]] = 1;
if(i%num[j] == 0)
break;
}
}
}
int main()
{
int n,q;
scanf("%d %d",&n,&q);
isPrime(n);
int k;
while(q--)
{
scanf("%d",&k);
printf("%lld\n",num[k-1]);
}
return 0;
}
T3 乘法逆元
题目链接：洛谷P3811
题目解析
需用到逆元递推公式：inv[i] = (p-p/i) * inv[p%i] % p;
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e7;
ll inv[MAXN];
int main()
{
ll n,p;
scanf("%lld %lld",&n,&p);
inv[1] = 1;
printf("1\n");
for(int i = 2; i <= n; i++)
{
inv[i] = (p-p/i) * inv[p%i] % p;
printf("%lld\n",inv[i]);
}
return 0;
}
T4 乘法逆元2
题目链接：洛谷P5431
题目解析
先通分后取模，每次k乘上k本身（k,k2,k3，不用快速幂），通分求和后分子乘上分母的逆元再取模即为结果。
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll p;
const int MAXN = 5e6+10;
ll num[MAXN];
ll fastpow(ll x,ll y)
{
x %= p;
ll res = 1;
while(y)
{
if(y&1)
res = res * x % p;
y >>= 1;
x = x * x % p;
}
return res;
}
ll inv(ll x)
{
return fastpow(x,p-2);
}
int main()
{
ll n,k;
scanf("%lld %lld %lld",&n,&p,&k);
ll t = k%p;
ll up = 0,down = 1;
for(ll i = 0; i < n; i++)
{
scanf("%lld",&num[i]);
up = (up*num[i]+down*t)%p;
down = down * num[i] % p;
t = t * k % p;
}
printf("%lld",up*inv(down)%p);
return 0;
}
T5 质数口袋
题目链接：洛谷P5723
题目解析
埃氏筛打表后每次从头遍历即可，注意打表数组要开bool型数组，不然会MLE。
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e5;
bool prime[MAXN];
void isPrime()
{
prime[0] = prime[1] = true;
for(ll i = 2; i*i <= MAXN; i++)
{
if(!prime[i])
for(ll j = i*i; j <= MAXN; j += i)
prime[j] = true;
}
}
int main()
{
int L;
scanf("%d",&L);
isPrime();
int sum = 0,cnt = 0;
for(int i = 2;; i++)
{
if(!prime[i])
{
if(sum + i <= L)
{
sum += i;
cnt++;
printf("%d\n",i);
}
else
break;
}
}
printf("%d",cnt);
return 0;
}
T6 回文质数
题目链接：洛谷P1217
题目解析
这道题我是用提前打表写的，打表后直接查询遍历输出即可。
打表代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 100000010;
bool prime[MAXN];
void isPrime()
{
prime[0] = prime[1] = true;
for(ll i = 2;i*i <= MAXN;i++)
{
if(!prime[i])
for(ll j = i*i;j <= MAXN;j += i)
prime[j] = true;
}
}
ll reverse( ll number )
{
ll num = 0,flag = 0;
if(number < 0)
{
number = -number;
flag = 1;
}
while(number)
{
num += number % 10;
if((number > 10) || (number % 10 == 0))
num *= 10;
number /= 10;
}
if(flag)
num = -num;
return num;
}
int main()
{
FILE *fp;
fp = fopen("1.txt","w");
isPrime();
for(ll i = 0;i <= MAXN;i++)
if(!prime[i])
{
if(i == reverse(i))
fprintf(fp,"%lld,",i);
}
fclose(fp);
return 0;
}
AC代码
#include<bits/stdc++.h>
using namespace std;
//typedef long long ll;
int prime[] = {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;
scanf("%d %d",&a,&b);
for(auto i:prime)
if(i >= a && i <= b)
printf("%d\n",i);
else if(i > b)
break;
return 0;
}
T7 哥德巴赫猜想（升级版）
题目链接：洛谷P1579
题目解析
埃氏筛打表后双重循环找出符合条件的一组输出即可。
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e6+10;
int prime[MAXN];
void isPrime()
{
prime[0] = prime[1] = 1;
for (ll i = 2; i * i < MAXN; i++)
{
if (!prime[i])
{
for (ll j = i * i; j < MAXN; j += i)
prime[j] = 1;
}
}
}
int main()
{
isPrime();
int n;
scanf("%d", &n);
for(int i = 2; i < n-1; i++)
for(int j = 2; j < n-1-i; j++)
{
int k = n-i-j;
if(prime[i] == 0 && prime[j] == 0 && prime[k] == 0)
if(i+j+k == n)
{
printf("%d %d %d", i, j, k);
return 0;
}
}
return 0;
}
T8 A % B Problem
题目链接：洛谷P1865
题目解析
埃氏筛打表后每次在区间内遍历即可，注意判断区间是否成立。
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e6+10;
bool prime[MAXN];
void isPrime()
{
prime[0] = prime[1] = true;
for(ll i = 2;i*i <= MAXN;i++)
{
if(!prime[i])
for(ll j = i*i;j <= MAXN;j += i)
prime[j] = true;
}
}
int main()
{
int n,m;
int l,r;
scanf("%d %d",&n,&m);
isPrime();
while(n--)
{
scanf("%d %d",&l,&r);
if(l >= 1 && l <= m && r >= 1 && r <= m && l <= r)
{
int cnt = 0;
for(int i = l;i <= r;i++)
if(!prime[i])
cnt++;
printf("%d\n",cnt);
}
else
printf("Crossing the line\n");
}
return 0;
}
T9 素数个数
题目链接：洛谷P3912
题目解析
埃氏筛打表时统计素数个数，打表完成后直接输出个数即可。
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e8+10;
bool prime[MAXN];
void isPrime(ll n)
{
ll cnt = 0;
prime[0] = prime[1] = 1;
for(ll i = 2;i <= n;i++)
{
if(!prime[i])
{
cnt++;
for(ll j = i*i;j <= n;j += i)
prime[j] = true;
}
}
printf("%lld",cnt);
}
int main()
{
ll n;
scanf("%lld",&n);
isPrime(n);
return 0;
}
T10 因子和
题目链接：洛谷P1593
题目解析
逐渐看不懂的题解
分解质因数并记录每个质因数的个数（记为幂数一起存储），结果为所有质因数等比数列求和再相乘。
（要注意特判逆元为0的情况）
AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 9901,MAXN = 1e6;
ll num[MAXN][2],sum[MAXN];
ll fastpow(ll x,ll y)
{
x %= mod;
ll res = 1;
while(y)
{
if(y&1)
res = res * x % mod;
y >>= 1;
x = x * x % mod;
}
return res;
}
ll inv(ll x)
{
return fastpow(x,mod-2)%mod;
}
int main()
{
ll a,b;
scanf("%lld %lld",&a,&b);
ll t = a;
ll cnt = 0;
for(ll i = 2;i*i <= a;i++)
{
if(t%i == 0)
num[++cnt][0] = i;
while(t%i == 0)
{
num[cnt][1]++;
t /= i;
}
}
if(t != 1)
{
num[++cnt][0] = t;
num[cnt][1] = 1;
}
ll res = 1;
for(ll i = 1;i <= cnt;i++)
{
sum[i] = (fastpow(num[i][0],num[i][1]*b+1)-1)*inv(num[i][0]-1)%mod;
if(sum[i] == 0)
sum[i] = num[i][1]*b+1;
res = res * sum[i] % mod;
}
printf("%lld",(res%mod+mod)%mod);
return 0;
}
