题意:求区间内出现过的奇数是偶数,出现过的偶数是奇数的个数。
析:这个题是要三进制进行操作的。dp[i][j] 表示前 i 位,状态是 j,可以用三进制来表示 0表示没有出现,1表示奇数,2表示偶数。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){ return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[22][60000];
int a[22], b[12], f[12];
void g(int val){
for(int i = 9; i >= 0; --i){
b[i] = val / f[i];
val %= f[i];
}
}
bool judge(int val){
g(val);
for(int i = 0; i < 10; ++i)
if((i&1) && b[i] == 1) return false;
else if(!(i&1) && b[i] == 2) return false;
return true;
}
int cal(int num, int val){
g(val);
b[num] = (b[num]&1) ? 2 : 1;
val = 0;
for(int i = 0; i < 10; ++i) val += b[i] * f[i];
return val;
}
LL dfs(int pos, int val, bool is, bool ok){
if(!pos) return judge(val);
LL &ans = dp[pos][val];
if(!ok && ans >= 0) return ans;
LL res = 0;
int n = ok ? a[pos] : 9;
for(int i = 0; i <= n; ++i)
if(is && !i) res += dfs(pos-1, val, is, ok && i == n);
else res += dfs(pos-1, cal(i, val), false, ok && i == n);
return ok ? res : ans = res;
}
LL solve(LL n){
int len = 0;
while(n){
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 0, true, true);
}
int main(){
f[0] = 1;
for(int i = 1; i < 10; ++i) f[i] = f[i-1] * 3;
memset(dp, -1, sizeof dp);
int T; cin >> T;
while(T--){
LL m, n;
cin >> m >> n;
cout << solve(n) - solve(m-1) << endl;
}
return 0;
}
原文:http://www.cnblogs.com/dwtfukgv/p/5958472.html