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数位DP貌似很弱,先切水的练练手~ HDU2089:点击打开链接 输出区间里面没有连续62,没有4的数字个数. #include using namespace std; long long n, m; int bit[11], l; int dp[11][2]; int dfs (int pos, bool f1, bool f2) {//当前的位数 前一个数是不是6 是不是可以取到

吉哥系列故事——恨7不成妻 Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 2758    Accepted Submission(s): 829 Problem Description 　　单身! 　　依然单身！

F(x) Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3315    Accepted Submission(s): 1247 Problem Description For a decimal number x

Good NumbersTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3814    Accepted Submission(s): 1213Problem DescriptionIf we sum up every digit of a

Problem Description A wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- string "13" and can be divided by 13. For example, 130 and 2613 are wqb-numbe

连接:点击打开链接 题意:定义g(x)为f(1)到f(n)中模k余n的个数,求出所有g(x)的异或和. 坑爹UVA一直报RE然后转向hihocoder就过了~ f(2x)=3f(x),f(2x+1)=f(2x)+1. 然后发现f(1)到f(n)在三进制下等价于二进制下的1到n,然后就可以用数位DP搞了. 这个思想下的转移方程dp(d,m) = dp(d-1, (m-x)%k). #in

K-th Nya Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others) Total Submission(s): 2762    Accepted Submission(s): 869 Problem Description Arcueid li