hdu 5106 Bits Problem(数位dp)
2014-11-16 11:13
423 查看
题目链接:hdu 5106 Bits Problem
题目大意:给定n和r,要求算出[0,r)之间所有n-onebit数的和。
解题思路:数位dp,一个ct表示个数,dp表示和,然后就剩下普通的数位dp了。不过貌似正解是o(n)的算法,但是n才
1000,用o(n^2)的复杂度也是够的。
题目大意:给定n和r,要求算出[0,r)之间所有n-onebit数的和。
解题思路:数位dp,一个ct表示个数,dp表示和,然后就剩下普通的数位dp了。不过貌似正解是o(n)的算法,但是n才
1000,用o(n^2)的复杂度也是够的。
#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; }
相关文章推荐
- HDU 5106 Bits Problem(数位DP)
- HDU 5106 Bits Problem(数位DP->二进制大数模拟)@
- [数位dp] hdu 5106 Bits Problem
- 【HDU】5106 Bits Problem 数位DP
- HDU 5106 Bits Problem (数位DP)
- hdu 5106 Bits Problem(数位dp)
- 【数位DP】 HDOJ 5106 Bits Problem
- hdu 5106 hdu 5435 数位dp
- HDU 3271 SNIBB [数位DP]
- HDU-3555 Bomb 数位DP
- HDU 3555 Bomb (数位DP)
- HDU 4507 数位dp
- hdu 3652(数位dp)
- hdu 3555 数位dp
- HDU 3652 3555 数位dp
- HDU 3652 数位DP
- HDU-3555 Bomb 数位DP
- HDU 4389 X mod f(x) [数位DP]
- HDU 3555 Bomb(数位DP)
- HDU 4389 X mod f(x)[数位统计dp]