acdream 1726 A Math game (部分和問題 DFS剪枝)
Time Limit: 2000/1000MS (Java/Others)
Memory Limit: 256000/128000KB (Java/Others)
Problem Description
Recently, Losanto find an interesting Math game. The rule is simple: Tell you a number
H, and you can choose some numbers from a set {a[1],a[2],......,a[n]}.If the sum of the number you choose is
H, then you win. Losanto just want to know whether he can win the game.
Input
There are several cases.
In each case, there are two numbers in the first line
n (the size of the set) and
H. The second line has n numbers {a[1],a[2],......,a[n]}.
0,All the numbers areintegers.
Output
If Losanto could win the game, output "Yes" in a line. Else output "No" in a line.
Sample Input
10 87
2 3 4 5 7 9 10 11 12 13
10 38
2 3 4 5 7 9 10 11 12 13
Sample Output
No
Yes
Source
第九屆北京化工大學程序設計競賽
題目鏈接：http://acdream.info/problem?pid=1726
題目大意：部分和問題
題目分析：正解是二分＋DFS，我的做法是DFS用前綴和剪枝，4ms過，估計還是數據水了，
#include
#include
#define ll long long
ll a[45], h, sum[45];
int n;
bool flag;
void DFS(ll num, int pos)
{
if(num == h)
{
flag = true;
return;
}
if(flag || pos == n + 1)
return;
//若當前數字加剩下的數字和小於h或者當前數字大於h則返回
//不加妥T
if(sum[n] - sum[pos - 1] + num < h || num > h)
return;
DFS(num + a[pos], pos + 1);
DFS(num, pos + 1);
return;
}
int main()
{
while(scanf("%d %lld", &n, &h) != EOF)
{
memset(sum, 0, sizeof(sum));
flag = false;
for(int i = 1; i <= n; i++)
{
scanf("%lld", &a[i]);
sum[i] = sum[i - 1] + a[i];
}
DFS(0, 1);
if(flag)
printf("Yes\n");
else
printf("No\n");
}
}
