题目链接 脑子有点乱,有的地方写错了,尚大婶鄙视了... 来个模版的. #include <iostream> #include <cstdio> #include <cstring> using namespace std; #define LL __int64 LL dp[][]; ]; LL dfs(int pos,int pre,int bound) { int end,tpre,i; LL ans = ; ) ; ) return dp[pos][pre];…

原题直通车: HDU  4722  Good Numbers 题意: 求区间[a,b]中各位数和mod 10==0的个数. 代码: #include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; int f[20]; long long work(long long x){ long long ret=0, u=x; int t=0, s=0…

HDOJ(HDU).1058 Humble Numbers (DP) 点我挑战题目 题意分析 水 代码总览 /* Title:HDOJ.1058 Author:pengwill Date:2017-2-15 */ #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #define nmax 6000 #define min(a,b) (a<b?a:b…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4722 Good Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 422    Accepted Submission(s): 146 Problem Description If we sum up every digit o…

Description If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number.  You are required to count the number of good numbers in the range from A to B, inclusive.   Input The first line has a…

Good Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3467    Accepted Submission(s): 1099 Problem Description If we sum up every digit of a number and the result can be exactly divided b…

If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number. You are required to count the number of good numbers in the range from A to B, inclusive. InputThe first line has a number T (T <=…

数位DP!!! 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector> #define ll __int64 #define pi acos(-1.0) #define MAX 50000 using namespac…

类型:数位DP 题意:定义一个Good Number 为 一个数所有位数相加的和%10==0.问[A,B]之间有多少Good Number. 方法: 正常“暴力”的定义状态:(i,d,相关量) 定义dp[i][d][mod] 为 d开头的i位数中,%10==mod的数的个数 dp[i][d][mod] = sum(dp[i-1][0~9][(mod-d+10)%10] 出口:dp[1][d][mod] = (d==mod); 代码: #include <cstdio> #include <…

#include<iostream> #include<cstring> #include<cstdio> #include<vector> #include<queue> using namespace std; #define ll long long #define maxn 100050 int ok(ll n){ *;i<=n;i++){ ll sum = ,tmp = i; while(tmp){ sum += tmp%; tm…