HDU 3555 数位DP

最新推荐文章于 2018-10-23 16:48:58 发布

明太鱼

最新推荐文章于 2018-10-23 16:48:58 发布

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分类专栏：动态规划

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本文详细介绍了使用C++编程语言解决复杂数学问题的方法，包括数学库的应用、算法优化和错误处理技巧。

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
#include <cassert>
#include <complex>
using namespace std;
typedef long long ll;
typedef long double ld;
const int int_max = 0x07777777;
const int int_min = 0x80000000;
const int inf=0x20202020;
const ll mod=1000000007;
const double eps=1e-9;
const double pi=3.1415926535897932384626;
const int DX[]={1,0,-1,0},DY[]={0,1,0,-1};
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
bool cmp (const void *a , const void *b )
{
    return *(int *)a < *(int *)b;
}
ll dp[30][3];
ll n,a[70];
ll solve (int len){
    ll sum = 0,last = 0;
    int flag = 0;
    for(int i = len; i >= 1; i--){
        sum += dp[i-1][1]*a[i];
        if(flag) sum += dp[i-1][0]*a[i];
        if(!flag && a[i]>4) sum += dp[i-1][2];
        if(last==4 && a[i]==9) flag = 1;
        last = a[i];
    }
    return sum;
}
int main()
{
    for(int i = 0; i <30; i++){
        for(int j = 0; j < 3; j++) dp[i][j] = 0;
    }
    dp[0][0] = 1;
    for(int i = 1; i < 30; i++){
        dp[i][0] = dp[i-1][0]*10 - dp[i-1][2];
        dp[i][1] = dp[i-1][1]*10 + dp[i-1][2];
        dp[i][2] = dp[i-1][0];
    }
    int T;
    while(scanf("%d", &T)!=EOF){
        while(T--){
            scanf("%I64d", &n);
            n++;
            for(int i = 0; i < 30; i++) a[i] = 0;
            int len = 0;
            while(n){
                a[++len] = n%10;
                n /= 10;
            }
            ll ret = solve(len);
            printf("%I64d\n",ret);
        }
    }
}