HDU

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 <= 10000) , indicating the number of test cases.
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 10 ¹⁸).OutputFor test case X, output "Case #X: " first, then output the number of good numbers in a single line.Sample Input

2
1 10
1 20

Sample Output

Case #1: 0
Case #2: 1

        
 

Hint

The answer maybe very large, we recommend you to use long long instead of int.

        
 这题有两种做法，找规律或者数位dp
法一找规律：从0开始打表会发现每10个数都有一个good number 0 - 9， 10- 19 …… 这样假如要求0 到123，只需要求 0 - 119的 有12个good numbers，再暴力求120 - 123的即可。
法二数位dp：这道题在数位dp中算是模板入门了吧。

法一代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <cstring>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
char st[12];
ll x;
int main()
{
    int t;
    ll a, b;
    scanf("%d", &t);
    ll cnta = 0, cntb = 0;
    for(int cas = 1; cas <= t; ++cas) {
        cnta = 0;
        cntb = 0;
        scanf("%lld %lld", &a, &b);
        a--;
        if(a < 0) cnta--;
        a = max(a, ll(0));
        ll ma = a % 10;
        cnta += a / 10;
        for(ll i = 0; i <= ma; ++i) {
            ll sum = 0;
            ll v = a / 10 * 10+ i;
            while(v) {
                sum += v % 10;
                v /= 10;
            }
            if(sum % 10 == 0) cnta++;
        }
        ll mb = b % 10;
        cntb += b / 10;
        for(ll i = 0; i <= mb; ++i) {
            ll sum = 0;
            ll v = b / 10 * 10+ i;
            while(v) {
                sum += v % 10;
                v /= 10;
            }
            if(sum % 10 == 0) cntb++;
        }
        
        printf("Case #%d: %lld
", cas, cntb - cnta);
    }
    return 0;
}

法二代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <cstring>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
ll dp[22][188];
ll ed[22];
ll dfs(int pos, ll sum, bool lmt) {
    if(pos == 0) {
        if(sum % 10 == 0) return 1;
        return 0;
    }
    
    if(!lmt && dp[pos][sum] != -1) return dp[pos][sum];
    ll ans = 0;
    ll up =  lmt? ed[pos] : 9;
    for(ll i = 0; i <= up; ++i) {
        ans += dfs(pos - 1, sum + i, lmt && i == ed[pos]);
    }
    if(!lmt) dp[pos][sum] = ans;//统计状态
    return ans;
}
ll solv(ll x) {
    if(x < 0) return 0;
    int len = 0;
    while(x) {
        ed[++len] = x % 10;
        x /= 10;
    }
    
    return dfs(len, 0, 1);
}

int main()
{
    int t;
    ll a, b;
    scanf("%d", &t);
    memset(dp,-1,sizeof(dp));
    for(int cas = 1; cas <= t; ++cas) {
        
        scanf("%lld %lld", &a, &b);
        printf("Case #%d: %lld
", cas, solv(b) - solv(a - 1));
    }
    return 0;
}