JZOJ 5184. 【NOIP2017提高组模拟6.29】Gift

5184. 【NOIP2017提高组模拟6.29】Gift (Standard IO)

Time Limits:  1000 ms  Memory Limits: 262144 KB  Detailed Limits  

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Description

 

Input

Output

 

Sample Input

input 1:
6 25
8 9 8 7 16 5
input 2:
30 250
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Sample Output

output 1:
15
output 2:
6509431

 

Data Constraint

 

做法：设f[i][j]表示取i到n个物品，花费为j的方案数，统计答案时，假设此时无法选取的最小物品为x，除去必须要选的物品剩余Val，那么答案为f[x + 1][Val - c[x] + 1 ~ Val]

 

代码如下：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 #define mo 10000007
 6 #define LL long long
 7 #define N 1007
 8 using namespace std;
 9 int n, m, a[N], Pre[N];
10 LL f[N][N], ans;
11 
12 int main()
13 {
14     scanf("%d%d", &n, &m);
15     for (int i = 1; i <= n; scanf("%d", &a[i++]));
16     sort(a + 1, a + n + 1);
17     for (int i = 1; i <= n; i++)
18         Pre[i] = Pre[i - 1] + a[i];
19     if (Pre[n] <= m)
20     {
21         printf("1");
22         return 0;
23     }
24     f[n + 1][m] = 1;
25     for (int i = n; i; i--)
26     {
27         for (int j = Pre[i - 1], k = min(m + 1, Pre[i - 1] + a[i]); j < k; j++)    ans = (ans + f[i + 1][j]) % mo;
28         for (int j = m - a[i]; j >= 0; j--)
29             f[i][j] = (f[i][j] + f[i + 1][j] + f[i + 1][j + a[i]]) % mo;
30         for (int j = m; j >= m - a[i] + 1; j--)
31             f[i][j] = (f[i][j] + f[i + 1][j]) % mo;
32     }
33     printf("%lld", ans);
34 }

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