UVA 12486 Space Elevator（数位DP）

题目pdf：http://acm.bnu.edu.cn/v3/external/124/12486.pdf

大致题意：求第n个不包含"4"和"13"为子串的数是多少 ， n<= 1e18

思路：就是一般的数位DP，二分答案，对答案的数求数位DP算出此数以内有多少个满足条件的数

但是....居然答案爆long long，要用unsigned long long 才能过，就这个坑点

// 0 ms 0 KB 2633 B 2015-08-15 01:02:36 
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define reveach(i, v) for (__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
#define rep(i,n) for ( int i=0; i< int(n); i++ )
using namespace std;
typedef unsigned long long ull;
#define X first
#define Y second
typedef pair<int,int> pii;

template <class T>
inline bool RD(T &ret) {
    char c; int sgn;
    if (c = getchar(), c == EOF) return 0;
    while (c != '-' && (c<'0' || c>'9')) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return 1;
}
template <class T>
inline void PT(T x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9) PT(x / 10);
    putchar(x % 10 + '0');
}

ull dp[20][3];
/*
 dp[i][0] 不包含4和13
 dp[i][1] 不包含4和13最高位为3
 dp[i][2] 包含4或13
 */
int bit[20];
ull cal(ull x){
        ull t = x;
        int c = 0;
        while(t){
                bit[++c] = t%10;
                t /= 10;
        }
        bit[c+1] = 0;
        ull ans = 0;
        bool is_13 = 0;
        for(int p = c; p >= 1; p--){
                ans += bit[p]*dp[p-1][2];
                if(is_13) ans += bit[p]*dp[p-1][0];
                else{
                        if(bit[p] > 4) ans += dp[p-1][0];
                        if(bit[p] > 1) ans += dp[p-1][1];
                        if(bit[p+1] == 1 && bit[p] > 3) ans += dp[p-1][0];
                        if( (bit[p+1] == 1 && bit[p] == 3)||bit[p] == 4)is_13 = 1;
                }
        }
        return x-ans;
}
int main(){
        dp[0][0] = 1;
        REP(i,18){
                dp[i][0] = dp[i-1][0]*9-dp[i-1][1];
                dp[i][1] = dp[i-1][0];
                dp[i][2] = dp[i-1][2]*10+dp[i-1][0]+dp[i-1][1];
        }


        ull n;
        while(RD(n)){
                ull lb = 0,ub = (ull)1000000000000000000*18;
                while(ub-lb > 1){
                        ull mid = lb/2 + ub/2 + ((lb&1)+(ub&1))/2;
                        if(cal(mid) <= n) lb = mid;
                        else ub = mid;
                }
                PT(lb);puts("");
        }

}