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…

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…

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,…

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…

[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…

递推,但是要注意细节.题目的意思,就是求s(x) = i+(i+1)+(i+2),i<n.该表达中计算过程中CA恒为0(包括中间值)的情况.根据所求可推得.1-10: 31-100: 3*41-1000: 3*4*41-10000: 3*4*4*41-10^n: 3*4^(n-1).并且需要注意,一旦发现某一位大于3,则应立即跳出累加的循环.比如,f(133) = 24,f(143) = 24.同时,单独讨论个位的情况.28行的break处理该种情况. #include <cstdio>…

//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());…

题目大意:有一个关于 简单加法表达式  的定义告诉你,就是  选一个数字i  如果 i+(i+1)+(i+2) 它的和,没有任何一位进位的话,那就是 一个i的简单加法表达式,求小于n的表达式数目. 题解:排列组合分类讨论即可…… #include <cstdio> #include <cmath> #include <cstring> using namespace std; char s[15]; int solve(int i,int p){ if(p==1)ret…

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=31329#problem/V 使用题目所给函数,单单从某一个数字来看,就是直接求这个数各个数位上的和:而且p=====>q之间的数调用这个函数,其数值都是在1~9之间:因此,求x和y%10的值,然后就直接45*((y-x)/10 );45是1+2+...+9的和,后面代表,p和q之间拥有多少个满足条件的组数,然后直接使用DFS递推即可 #include<map> #include…