题目链接:uva 10994 - Simple Addition 题目大意:给出l和r,求∑(l≤i≤r)F(i), F(i)函数题目中有. 解题思路:由两边向中间缩进,然后l和r之间的数可以按照1~9划分(只会有这几种情况). #include <stdio.h> #define ll long long ll ans; ll f(ll x) { if (x == 0) return 0; else if (x % 10) return x % 10; else return f(x / 1…

//组合数学 //计算sum{i从右往左数的第一个非0数字,p<=i<=q}. #include <cstdio> typedef long long ll; ll sum(ll n) { ll ans = , x; while(n) { x = n % ; n /= ; ans += (( + x) * x) / + n * ; //当个位在循环的时候,高位的朋友你在干嘛? } return ans; } int main() { ll a, b; ) { printf());…

Problem E Simple Addition Input: Standard Input Output: Standard Output Let’s define a simple recursive function F (n), where Let’s define another function S (p, q), In this problem you have to Calculate S (p, q) on given value of   p and q.   Input…

hdu 2451 Simple Addition Expression Problem Description A luxury yacht with 100 passengers on board is sailing on the sea in the twilight. The yacht is ablaze with lights and there comes out laughers and singing from the hall where an evening party i…

题目链接:uva 12253 - Simple Encryption 题目大意:给定K1.求一个12位的K2,使得KK21=K2%1012 解题思路:按位枚举,不且借用用高速幂取模推断结果. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; const ll ite=(1<<20)-1; ll N; /* l…

主题链接:http://acm.hdu.edu.cn/showproblem.php?pid=2451 Problem Description A luxury yacht with 100 passengers on board is sailing on the sea in the twilight. The yacht is ablaze with lights and there comes out laughers and singing from the hall where an…

Problem Description A luxury yacht with 100 passengers on board is sailing on the sea in the twilight. The yacht is ablaze with lights and there comes out laughers and singing from the hall where an evening party is in full swing. People are singing,…

UVA - 10014 Simple calculations Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu id=19100" style="color:blue">Submit Status Description  Simple calculations  id=19100" style="color:blue">The Pro…

[来源] 2008年哈尔滨区域赛 [题目链接]: http://acm.hdu.edu.cn/showproblem.php?pid=2451 [参考博客]: HDU 2451 Simple Addition Expression [题意]: 题意是要判断前n位数字(不包括n),有多少个数字 i 跟前面两个 i+1 , i+2 ,相加时不进位 . 符合要求的数字就是个位 0 ~ 2 ,其余位 0 ~ 3. 用一个dfs就可以搜出来了. 对于当前位是 x 的话 , 若果 x > 3 , 可以直接得…

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=955 根据递推公式推倒出a1的公式. a1=(n*a0+an+1-2*(n*c1+(n-1)*c2+...+cn))/(n+1); #include <cstdio> #include <cstring> #include <algorithm> #defi…