hdu 5435 A serious math problem

A serious math problem

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Xiao Jun likes math and he has a serious math question for you to finish.
Define (F[x]) to the (xor) sum of all digits of (x) under the decimal system,for example (F(1234) = 1 xor 2 xor 3 xor 4 = 4).
Two numbers (a,b(a≤b)) are given,figure out the answer of (F[a] + F[a+1] + F[a+2]+…+ F[b−2] + F[b−1] + F[b]) doing a modulo (10^9+7).

Input

The first line of the input is a single integer (T(T<26)), indicating the number of testcases.
Then (T) testcases follow.In each testcase print three lines :
The first line contains one integers (a).
The second line contains one integers (b).
(1≤|a|,|b|≤100001),(|a|) means the length of (a).

Output

For each test case, output one line "Case #x: y", where x is the case number (starting from 1) and y is the answer.

Sample Input

4
0
1
2
2
1
10
9999
99999

Sample Output

Case #1: 1
Case #2: 2
Case #3: 46
Case #4: 649032

Source

BestCoder Round #55 ($)

这就是数位dp。。。。又想起了当年我作为一个被毒害的小朋友去写windy数的恐惧（哪天再去做一遍233）
我们来简单说一下这道题的做法：
(dp[i][j]) 表示(i)位数，异或值为(j)的数的个数。((e.g. 0-9 10 - 99 100 - 999 ......))
表示这个转移很暴力~~（代码展示---）
然而数位dp和别的不同的是，dp其实只是你的预处理。。。。你可能还要计数。。。（蒟蒻表示计数比dp难）
需要我们做的操作是求出所有小于等于(a)的自然数的异或值的和。而我们预处理的是很多个区间的答案，所以我们还要统计一下。
我们需要把(a)这个数给拆成很多个区间。。。
举个栗子：
(3122=(0~999)+(1000~1999) + (2000~2999) + (3000 ~ 3589))
((3000~3122)=(3000~3099)+(3100~3122))
((3100~3122)=(3100~3109) + (3110~3119) + (3120~3122))
按这种方法分解以后，我们就和计数啦~
（注意：代码中可以看出，我们传进去是(a)，实际上返回的是(a-1)的答案，所以适当调整一下~）

#include<bits/stdc++.h>
using namespace std;
const int maxn = 100005, mod = 1e9 + 7;
char a[maxn], b[maxn];
int dp[maxn][20], ans[20];
int cas, last, qwe;

inline void prepare()
{
	dp[0][0] = 1;
	for(int i = 0; i <= 9; ++i) dp[1][i] = 1;
	for(int i = 2; i < maxn; ++i)
		for(int j = 0; j < 10; ++j)
			for(int k = 0; k <= 15; ++k)
				dp[i][j ^ k] += dp[i - 1][k], dp[i][j ^ k] %= mod; 
}

inline int workk(char * str)
{
	int len = strlen(str + 1); int pre = 0; int ret = 0;
	memset(ans, 0, sizeof(ans));
	for(int i = 1; i <= len; ++i)
	{
		int num = str[i] - '0';
		for(int j = 0; j < num; ++j)
		{
			for(int k = 0; k <= 15; ++k)
			{
				ans[pre ^ j ^ k] += dp[len - i][k];
				ans[pre ^ j ^ k] %= mod;
			}
		}
		pre ^= num;
	}
	for(int i = 1; i <= 15; ++i)	ret = (ret + (long long)i * ans[i]) % mod;
	return ret; 
}

int main()
{
	prepare();
	int t; scanf("%d", &t);
	while(t--)
	{
		cas++; qwe = 0; last = 0;
		scanf("%s", a + 1); scanf("%s", b + 1); int p = strlen(b + 1);
		for (int i = 1; i <= p; i++)  last ^= (b[i] - '0'); 
		qwe = (workk(b) - workk(a) + mod) % mod;	
		qwe = (qwe + last) % mod;  
        printf("Case #%d: %d
", cas, qwe);  	
	}
	return 0;
}