HDU 5898 - odd-even number

Problem Description
For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even number.Now we want to know the amount of odd-even number between L,R(1<=L<=R<= 9*10^18).

Input
First line a t,then t cases.every line contains two integers L and R.

Output
Print the output for each case on one line in the format as shown below.

Sample Input

2
1 100
110 220


Sample Output

Case #1: 29

Case #2: 36

题意：奇偶数就是偶数的连续个数有奇数个，奇数的连续个数有偶数个，给出一个区间，求这个区间总共有多少个odd-even number。

一道数位DP题，用ans[i][j][k]表示前 i 位，前一位是 j，奇数或偶数有 k。具体注释看代码。

#include <cstdio>long long ans[20][11][20];//第i位，前一位是j，奇数或偶数有kint num[20];//把数字拆开//lead表示有无前导0//limit表示是否是上限long long solve(int pos, int state, int n, bool lead, bool limit){    //只有一个数的情况    if (pos == -1 && (state + n) % 2 == 1)        return 1;    if (pos == -1 && (state + n) % 2 == 0)        return 0;    if (!lead && !limit && ans[pos][state][n] != -1)        return ans[pos][state][n];    int up = limit ? num[pos] : 9;    long long ret = 0;    if (lead)    {        for (int i = 0; i <= up; ++i)            ret += solve(pos-1, i, 1, lead && i == 0, limit && i == num[pos]);    }    else if ((state + n) % 2 == 1)//奇数    {        for (int i = 0; i <= up; ++i)        {            if ((i + state) % 2 == 0)//奇数后是奇数                ret += solve(pos-1, i, n+1, lead && i == 0, limit && i == num[pos]);            else//奇数后是偶数                ret += solve(pos-1, i, 1, lead && i == 0, limit && i == num[pos]);        }    }    else    {        for (int i = 0; i <= up; ++i)        {            if ((i + state) % 2 == 0)                ret += solve(pos-1, i, n+1, lead && i == 0, limit && i == num[pos]);        }    }    if (!limit && !lead)        ans[pos][state][n] = ret;    return ret;}int main(){    int T;    long long left, right;    for (int i = 0; i < 20; ++i)        for (int j = 0; j < 11; ++j)            for (int k = 0; k < 20; ++k)                ans[i][j][k] = -1;    scanf("%d", &T);    for (int icase = 1; icase <= T; ++icase)    {        long long ans_left, ans_right;        scanf("%lld %lld", &left, &right);        left--;        int pos = 0;        num[0] = 0;        while (left != 0)        {            num[pos++] = left % 10;            left /= 10;        }        //printf("left ok\n");        ans_left = solve(pos-1, 10, 1, true, true);        //printf("solve left ok\n");        pos = 0;        num[0] = 0;        while (right != 0)        {            num[pos++] = right % 10;            right /= 10;        }        //printf("right ok\n");        ans_right = solve(pos-1, 10, 1, true, true);        //printf("solve right ok\n");        printf("Case #%d: %lld\n", icase, ans_right - ans_left);    }    return 0;}

PS.看到odd-even number想起大仓明日香的《prime number~君と出会える日~》

也许回过头来再看樱花庄里发生的故事只会感觉自己那么幼稚、愚笨，但是唯一不变和让自己感觉当初幼稚愚笨的原因，则是在幼稚和愚笨中的努力汗水泪水。