hdu 5106 Bits Problem(數位dp)，hdu5106

hdu 5106 Bits Problem(數位dp)，hdu5106

題目連結：hdu 5106 Bits Problem

題目大意：給定n和r，要求算出[0,r)之間所有n-onebit數的和。

解題思路：數位dp，一個ct表示個數，dp表示和，然後就剩下普通的數位dp了。不過貌似正解是o(n)的演算法，但是n才

1000，用o(n^2)的複雜度也是夠的。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int mod = 1000000007;const int maxn = 1005;int n;char s[maxn];ll bit[maxn], dp[maxn][maxn], ct[maxn][maxn];int solve () {    memset(ct, 0, sizeof(ct));    memset(dp, 0, sizeof(dp));    int l = strlen(s), c = n;    ll sum = 0;    for (int i = 0; i < l; i++) {        for (int k = 0; k <= n; k++) {            if (ct[i][k] == 0)                continue;            for (int j = 0; j < 2; j++) {                if (j && k == 0)                    continue;                ct[i+1][k-j] = (ct[i+1][k-j] + ct[i][k]) % mod;                dp[i+1][k-j] = (dp[i+1][k-j] + dp[i][k] + bit[l-i-1] * j * ct[i][k]) % mod;            }        }        for (int j = 0; j < s[i]-'0'; j++) {            ct[i+1][c-j] = (ct[i+1][c-j] + 1) % mod;            dp[i+1][c-j] = (dp[i+1][c-j] + sum + bit[l-i-1] * j) % mod;        }        if (s[i] == '1') {            sum = (sum + bit[l-i-1]) % mod;            c--;        }    }    return dp[l][0];}int main () {    bit[0] = 1;    for (int i = 1; i <= 1000; i++)        bit[i] = bit[i-1] * 2 % mod;    while (scanf("%d%s", &n, s) == 2) {        printf("%d\n", solve());    }    return 0;}