URAL 1057 数位DP(递归)

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题目：给出一个区间，问这个区间内的数，转换成B进制后，刚好有B个1的的数有多少个。

递归

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int k, b, l, r;int dp[66][66][66];int a[66];int dfs(int len, int cnt, bool lim) {if(!len) return cnt == k;if(!lim && ~dp[len][cnt][b]) return dp[len][cnt][b];int m = lim ? a[len-1] : b-1;int i, ret = 0;for(i = 0; i <= m && i <= 1; i++)ret += dfs(len-1, cnt+(i==1),lim&&i==m);if(!lim) dp[len][cnt][b] = ret;return ret;}int gao(int n) {int len = 0;while(n) { a[len++] = n%b; n /= b;}return dfs(len, 0, 1);}int main() {memset(dp, -1, sizeof(dp));while(cin >> l >> r >> k >> b) {cout <<  gao(r) - gao(l-1) << endl;}return 0;}

非递归

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int l, r, k, b;int dp[35][35];void init() {int i, j;memset(dp, 0, sizeof(dp));dp[0][0] = 1;for(i = 1; i <= 34; i++)for(j = 0; j <= i; j++) {    dp[i][j] += dp[i-1][j];    if(j) dp[i][j] += dp[i-1][j-1];}}int gao(int n) {int a[35], len = 0;while(n) {a[len++] = n%b; n/=b;}int i, j;int cnt = 0, ans = 0;bool flag = 0;for(i = len-1; i >= 0; i--) {for(j = 0; j < a[i]; j++) {if(j > 1) break;if(cnt+j> k) continue;ans += dp[i][k-cnt-j];}if(a[i] == 1) cnt++;if(a[i] > 1) { flag = 1; break; }}if(cnt == k && !flag) ans++;return ans;}int main() {while(cin >> l >> r >> k >> b) {init();cout << gao(r) - gao(l-1) << endl;}return 0;}