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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("");
}
}版权声明:本文为博主原创文章,未经博主允许不得转载。
UVA 12486 Space Elevator(数位DP)
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原文地址:http://blog.csdn.net/kalilili/article/details/47674055