基本原理 公钥与私钥的产生 随机选择两个不同大质数 $p$ 和 $q$，计算 $N=p\times q$ 根据欧拉函数，求得 $\phi(N)=\phi(p)\phi(q)=(p-1)(q-1)$ 选择一个小于 $\phi(N)$ 的整数 $e$，使 $e$ 和 $\phi(N)$ 互质。并求得 $e$ 关于 $\phi(N)$ 的模反元素，命名为 $d$，有 $ed \equiv 1 \pmod{\phi(N)}$ 将 $p$ 和 $q$ 的记录销毁 此时，$(N,e)$ 是公钥，$(N,d)$ 是私钥。

[公式] 韦达定理 [公式] 用法 1 [公式] 2 [公式] / [公式] / [公式] / [公式]

级数 调和级数 调和级数的定义 调和级数是指这样一个数列： / [公式] 调和级数的性质 $\gamma$是欧拉常数，其值为 / [公式] / 明显的, $S(n)$为第$n$个调和数。 / 已知 / [公式] / 欧拉推导过调和级数有限多项式和的表达式为

Crypto 写博客时的本机python环境：python 3.10 x64 cry1-babyRSA 题目附件 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 #!/usr/bin/env python3 import libnum import gmpy2 def main(): m = "" p = libnum.generate_prime(1024) q = libnum.generate_prime(1024) n = p * q e1 = 65537 e2 = 1145141 m = libnum.s2n(m) c1 = pow(m, e1, n) c2 = pow(m, e2, n) print(f"n = {n}") print(f"e1 = {e1}") print(f"e2 = {e2}") print(f"c1 = {c1}") print(f"c2 = {c2}") if __name__ == "__main__": main() # ------------ output ------------- # n = 11609263367794994463117283145812710043177521810736993971752031031462916890183901184704668542746877577916588155978013244385351397164066533771160861236441526284927774454246028029331726391203226023580325080150500633513024867014342350030181272221968801196510315424256352865890631054232306002238256568004250127485008008138279976475038656972273740968642332785779132654095393753232949667278798806004585797554024955342308244602767094536835410577382144435188162865642061122467384470501907391577779349252938141732012071206498806107556481558249549513041515803734342211746038126753951345855276903954190730328577080831957273691313 # e1 = 65537 # e2 = 1145141 # c1 = 8279258823057357102846768374381269167364145680055017957250521243478403606503599610855366519746944230676766499525422449675601214010991204564154995560170186683394412090168422510245266135032687364205431432451045158622417794414045719898864520112347836962316252383017549810699146506152781517871135246521405624365475969605452621085531890669372145482824845129281827033881675216546685064514926792907604133415349309151330709913454541960741984877203112442510747386406221828180805888471328964423290560512976977772551838742784356814497777401061881079781523967957560383718977490546677541952293716448514035557723329598904161762173 # c2 = 4995747575438050007737011353038705757162003396847797286289786278729187499823790079035532946676851313055563930519198963823829616599717198622635901839657079748022082189146477789049024407969208203999231434278100203042702919909473619456123328867313626560538182915794195719942071958092695261033449894563006040003298826647287929451919428024895476725340892133852628235964798488419924387986089462246202364608313134686465936926347848518960121189416319175083481701958106210362456062685045840587374473767109533027613795056920007028898921123363733374705988009798831764416119904696307107441325551226052940068337901039381485797771 解题思路 查看代码和注释掉的运行结果可知：m(密文)未知，已知e1,e2,c1,c2,n

stdin A+B / 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 #include <iostream> typedef long long ll; using std::cout; const char endl = '\n'; namespace sscio { inline ll pull() { ll sign = 1LL,res = 0LL; char ch = getchar(); for (;!isdigit(ch);ch=getchar()) { if (ch == '-') { sign *= -1LL; } } for (;isdigit(ch);ch = getchar()) { res = (res << 3) + (res << 1) - '0' + ch; // *10 } return sign * res; } } void solve() { ll n=sscio::pull(); ll p=sscio::pull(); cout << n + p << endl; } signed main() { solve(); return 0 ^ 0; }

【模板】线性筛素数 - 洛谷 / 【深基7.例2】质数筛 - 洛谷 / 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 #include <iostream> #include <bitset> typedef long long ll; const int sz=1e8+10; namespace sscio { inline ll pull() { ll sign = 1LL,res = 0LL; char ch = getchar(); for (;!isdigit(ch);ch=getchar()) { if (ch == '-') { sign *= -1LL; } } for (;isdigit(ch);ch = getchar()) { res = (res << 3) + (res << 1) - '0' + ch; // *10 } return sign * res; } } using std::bitset; bitset<sz> primeTable; int primes[sz],prpp=0; void fetchPrimes(const int &ed) { //线性筛 形参为边界 primeTable.reset(); primeTable.set(0); primeTable.set(1); for(register int cx=2;cx<ed;cx++) { if(!primeTable[cx]) { primes[prpp++]=cx; } for (register int cy=0;cy<prpp&&cx*primes[cy]<ed;cy++) { //用质数来筛质数 primeTable.set(cx*primes[cy]); if (cx%primes[cy]==0) { break; } } } } void solve() { fetchPrimes(sz-6); int n=sscio::pull(); //n为查询范围,因为已经筛了超出题目范围素数了,所以无需处理 int p=sscio::pull(); //查询的个数 while(p--) { int k=sscio::pull(); //表示查询第 k 小的素数 printf("%d\n", primes[k-1]); //因为我们下标从零开始 所以是k - 1 } } signed main() { solve(); return 0 ^ 0; } 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 #include <iostream> #include <bitset> typedef long long ll; const int sz=1e5+10; namespace sscio { inline ll pull() { ll sign = 1LL,res = 0LL; char ch = getchar(); for (;!isdigit(ch);ch=getchar()) { if (ch == '-') { sign *= -1LL; } } for (;isdigit(ch);ch = getchar()) { res = (res << 3) + (res << 1) - '0' + ch; // *10 } return sign * res; } } using std::bitset; bitset<sz> primeTable; int primes[sz],prpp=0; void fetchPrimes(const int &ed) { //线性筛 形参为边界 primeTable.reset(); primeTable.set(0); primeTable.set(1); for(register int cx=2;cx<ed;cx++) { if(!primeTable[cx]) { primes[prpp++]=cx; } for (register int cy=0;cy<prpp&&cx*primes[cy]<ed;cy++) { //用质数来筛质数 primeTable.set(cx*primes[cy]); if (cx%primes[cy]==0) { break; } } } } void solve() { fetchPrimes(sz-6); // int n=sscio::pull(); //n为查询范围,因为已经筛了超出题目范围素数了,所以无需处理 int p=sscio::pull(); //查询的个数 int cnt = 0; //空格数量 while(p--) { int k=sscio::pull(); if (!primeTable[k]) { if (cnt==0) { cnt++; } else { putchar(' '); } printf("%d", k); } } } signed main() { solve(); return 0 ^ 0; }

from Kotori 学长の小测验 / 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 from secret import flag from Crypto.Util.number import * # seed = bytes_to_long(flag) bits = seed.bit_length() while True: p = getPrime(bits + 1) if p > seed: break print(p) a = getRandomRange(1, p) b = getRandomRange(1, p) for _ in range(3): seed = (a * seed + b) % p print(seed) # 31893593182018727625473530765941216190921866039118147474754069955393226712079257707838327486268599271803 # 25820280412859586557218124484272275594433027771091486422152141535682739897353623931875432576083022273940 # 24295465524789348024814588142969609603624462580932512051939198335014954252359986260009296537423802567677 # 14963686422550871447791815183480974143372785034397446416396172429864269108509521776424254168481536292904 RSA的计算过程是：

一道AirDrop主题密码题wp / 题面 小A鼓起勇气向女神索要电话号码，但女神一定要考考他。女神说她最近刚看了一篇发表于安全顶会USENIX Security 2021的论文，论文发现苹果AirDrop隔空投送功能的漏洞，该漏洞可以向陌生人泄露AirDrop发起者或接收者的电话号码和电子邮箱。小A经过一番努力，获得了女神手机在AirDrop时传输的手机号哈希值，但再往下就不会了，你能继续帮助他吗？小A只记得女神手机号是170号段首批放号的联通号码。Hash：c22a563acc2a587afbfaaaa6d67bc6e628872b00bd7e998873881f7c6fdc62fc / flag格式：flag{13位电话号码（纯数字，含国家代码）}

AcWing 数组的部分 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 #include<cstdio> int main() { char t; scanf("%c",&t); double q[12][12]; // 初始化后全部读入 for (int i = 0;i < 12; i ++ ) for (int j = 0;j < 12; j ++ ) scanf("%lf",&q[i][j]); double s = 0, c = 0; for (int i = 1; i <= 5; i ++ ) for (int j = 0;j < #(i - 1)#；j ++ ) { s += q[i][j]; c += 1; } for (int i = 6; i <= 10 ;i ++ ) for (int j = 0; j <= #(10 - i)#; j ++ ) { s += q[i][j]; c += 1; } if ( t == 'S') printf ("%.1lf\n",s); else printf("%.1lf\n", s / c ); return 0; }

看题目中由 encode 可以猜到是一种编码方式 来源 / 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 <?php /* convert_uudecode() 函数对 uuencode 编码的字符串进行解码。 该函数常与 convert_uuencode() 函数一起使用。 */ $str = "Hello world!"; // 对字符串进行编码 $encodeString = convert_uuencode($str); echo $encodeString . "<br>"; // 对字符串进行解码(注意转义字符) $decodeString = convert_uudecode($encodeString); echo $decodeString; ?> 去这里运行试试>>>PHP在线运行