HDU 5185 Equation (linear dp full backpack), hdu5185
Equation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission (s): 64 Accepted Submission (s): 20
Problem DescriptionGorwin is very interested in equations. Nowadays she gets an equation like this
X1 + x2 + x3 + records + xn = n , And here 0 ≤ xi ≤ nfor1 ≤ I ≤ nxi ≤ xi + 1 ≤ xi + 1for1 ≤ I ≤ n −1
For a certain N , Gorwin wants to know how many combinations Xi Satisfies above condition.
For the answer may be very large, you are expected output the result after it modular M .
InputMulti test cases. The first line of the file is an integer T Indicates the number of test cases.
In the next T Lines, every line contain two integer N, m .
[Technical Specification]
1 ≤ T <20
1 ≤ n ≤50000
1 ≤ MB ≤ 1000000000
OutputFor each case output shoshould occupies one line, the output format is Case # id: ans, here id is the data number starting from 1, ans is the result you are expected to output.
See the samples for more details. Sample Input
23 1005 100
Sample Output
Case #1: 2Case #2: 3
Source BestCoder Round #32
Question link: http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 5185
Question: according to the formula given by the question, how many different methods can be used to obtain n, and the number of methods is equal to m?
Question Analysis: Because n is relatively large, it is not allowed to directly carry a backpack and time space. Considering the nature of the formula, n is obtained at the maximum, that is, 1 ~ Ma summation, ma * (ma + 1)/2 = n
To simplify the process, we can get ma = (sqrt (8n + 1)-1)/2. The time space complexity is reduced to nsqrt (n ), considering that dp [I] [j] represents the number of types of the first I number to synthesize the number j, the transfer equation is
Dp [I] [j] = dp [I-1] [j-I] + dp [I] [j-I], the number of types of j in the first numeric synthesis is equal to the sum of I and I in the j-I synthesis. dp [0] [0] = 1
#include <cstdio>#include <cmath>#include <algorithm>using namespace std;int dp[317][50001];int main(){ int T, n, m; scanf("%d", &T); for(int ca = 1; ca <= T; ca++) { dp[0][0] = 1; scanf("%d %d", &n, &m); int ans = 0, ma = (sqrt(8 * n + 1) - 1) / 2; for(int j = 1; j <= n; j++) for(int i = 1; i <= min(j, ma); i++) dp[i][j] = (dp[i][j - i] + dp[i - 1][j - i]) % m; for(int i = 1; i <= ma; i++) ans = (ans + dp[i][n]) % m; printf("Case #%d: %d\n", ca, ans); }}
