首页 > 代码库 > HDU 3709 Balanced Number (数位DP)

HDU 3709 Balanced Number (数位DP)

题意:找出区间内平衡数的个数,所谓的平衡数,就是以这个数字的某一位为支点,另外两边的数字大小乘以力矩之和相等,即为平衡数。

析:数位DP,dp[i][[j][k]表示 前 i 位以 j 为支点,还差 k 平衡,枚举 j 就好。

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>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e5 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};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[20][20][1500];int a[20];LL dfs(int pos, int dot, int balance, bool ok){    if(!pos)  return balance == 0;    if(balance < 0)  return 0;    LL &ans = dp[pos][dot][balance];    if(!ok && ans >= 0)  return ans;    LL res = 0;    int n = ok ? a[pos] : 9;    for(int i = 0; i <= n; ++i){        res += dfs(pos-1, dot, balance+(pos-dot)*i, ok && i == n);    }    if(!ok)  ans = res;    return res;}LL solve(LL n){    int len = 0;    while(n){        a[++len] = n % 10;        n /= 10;    }    LL ans = 0;    for(int i = 1; i <= len; ++i)        ans += dfs(len, i, 0, true);    return ans - len;//除去全是0的情况}int main(){    memset(dp, -1, sizeof dp);    LL x, y;    int T;  cin >> T;    while(T--){        scanf("%lld %lld", &x, &y);        printf("%lld\n", solve(y)-solve(x-1));    }    return 0;}

 

HDU 3709 Balanced Number (数位DP)