題目大意：

現(xiàn)在題目被加密了, 給出加密后的串

hjxh dwh v vxxpde,mmo ijzr yfcz hg pbzrxdvgqij rid stl mc zspm vfvuu vb uwu spmwzh.

一直前面4個詞是give you a number, 出題人說自己只會Fibonacci...

解密這一段文字然后寫程序

大致思路：

既然出題人說自己只會Fibonacci, 腦洞一下這個提議, 注意到前幾個字母： (h, g), (j, i), (x, v)..差距依次是1, 1, 2, 3, 5, 8....于是猜想字符差距是Fibonacci數(shù), 以26為循環(huán)節(jié)即可

得到解密之后的題面是：give you a number,and your task is calculating the sum of each digit in the number

于是就是個無聊的求按位之和的題了, 注意n是long long 范圍當(dāng)n取-2^63的時候轉(zhuǎn)正整數(shù)的long long會出錯就行了

代碼如下：

Result ?: ?Accepted ? ? Memory ?: ?1672 KB ? ? Time ?: ?0 ms

/* * this code is made by Gatevin * Problem: 1069 * Verdict: Accepted * Submission Date: 2015-09-13 22:10:16 * Time: 0MS * Memory: 1672KB */ /** Author: Gatevin* Created Time: 2015/9/12 12:12:47* File Name: Sakura_Chiyo.cpp*/ #include<iostream> #include<sstream> #include<fstream> #include<vector> #include<list> #include<deque> #include<queue> #include<stack> #include<map> #include<set> #include<bitset> #include<algorithm> #include<cstdio> #include<cstdlib> #include<cstring> #include<cctype> #include<cmath> #include<ctime> #include<iomanip> using namespace std; const double eps(1e-8); typedef long long lint; typedef unsigned long long ulint;/** hjxh dwh v vxxpde,mmo ijzr yfcz hg pbzrxdvgqij rid stl mc zspm vfvuu vb uwu spmwzh* give you a number* h - g = 1* j - i = 1* x - v = 2* h - e = 3* d - y = 26 + d - y = 5* w - o = 8* 猜想, 解密需要將所有字符加上Fibonacci數(shù)取模26得到, 前兩項是1 1*/int fib[100];int main() {//freopen("out.out", "w", stdout);fib[0] = fib[1] = 1;string s = "hjxh dwh v vxxpde,mmo ijzr yfcz hg pbzrxdvgqij rid stl mc zspm vfvuu vb uwu spmwzh";for(int i = 2, sz = s.length(); i < sz; i++)fib[i] = (fib[i - 1] + fib[i - 2]) % 26;int num = 0;for(int i = 0, sz = s.length(); i < sz; i++)if(s[i] != ' ' && s[i] != ',') s[i] = ((s[i] - 'a') - fib[num++] + 26) % 26 + 'a';//cout<<s<<endl;//s = "give you a number,and your task is calculating the sum of each digit in the number";lint n;while(scanf("%lld", &n) != EOF){ulint N;if(n < 0){N = (ulint)(-(n + 1)) + 1uLL;}else N = n;ulint ans = 0;while(N){ans += N % 10uLL;N /= 10uLL;}printf("%llu\n", ans);}return 0; }

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