HDU 3709 Balanced Number(数位dp)
2016-09-02 16:24
447 查看
题目链接:
HDU 3709 Balanced Number题意:
如果一个数字以某一位为平衡点左右力矩相等,则称该数字为Balanced Number。求区间[L,R]中Balanced Number的数量。数据范围:0≤L≤R≤1018
分析
枚举平衡点并记录平衡点左右力矩之差为sum,这样子才能记忆化。还要注意0的情况。Code
#include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> using namespace std; typedef long long ll; int digit[20]; ll dp[20][20][4000]; ll dfs(int pos, int pivot, int sum, int limit) { if (sum < 0 || sum > pivot * (pivot + 1) / 2 * 9) return 0; if (pos == -1) return sum == 0; if (!limit && dp[pos][pivot][sum] != -1) return dp[pos][pivot][sum]; int last = limit ? digit[pos] : 9; ll ret = 0; for (int i = 0; i <= last; ++i) { int next_sum = sum + i * (pos - pivot); ret += dfs(pos - 1, pivot, next_sum, limit && i == last); } if (!limit) dp[pos][pivot][sum] = ret; return ret; } ll solve(ll x) { if (x < 0) return 0; else if (x == 0) return 1; memset(digit, 0, sizeof(digit)); int len = 0; while (x) { digit[len++] = x % 10; x /= 10; } ll ret = 0; for (int i = 0; i < len; ++i) { ret += dfs(len - 1, i, 0, 1); } return ret - (len - 1); // 把0算了len次 } int main() { int T; ll L, R; scanf("%d", &T); while (T--) { memset(dp, -1, sizeof(dp)); scanf("%lld%lld", &L, &R); printf("%lld\n", solve(R) - solve(L - 1)); } return 0; }
相关文章推荐
- HDU 3709 Balanced Number ZOJ 3416 Balanced Number(数位DP)
- hdu 3709 Balanced Number(数位DP,5级)
- HDU 3709 Balanced Number (数位DP)
- HDU 3709 Balanced Number (简单数位DP)
- HDU 3709 Balanced Number(数位DP)
- HDU 3709 Balanced Number 枚举+数位DP
- HDU - 3709 Balanced Number (数位DP&记忆化dfs)好题
- hdu 3709 Balanced Number 数位dp
- HDU-3709 Balanced Number (数位DP)
- HDU 3709 Balanced Number (数位DP)
- HDU_3709 Balanced Number 数位dp
- hdu 3709 Balanced Number 数位DP
- HDU 3709 Balanced Number 数位dp
- HDU 3709 Balanced Number (数位dp)
- hdu 3709 Balanced Number(数位DP,5级)
- HDU 3709: Balanced Number (数位DP)
- HDU 3709 Balanced Number 数位DP
- 【数位DP】 HDU 3709 Balanced Number
- HDU 3709 Balanced Number(数位DP)
- 【HDU】3709 Balanced Number 数位DP