USACO 2.2 Subset Sums
For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical.
For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical:
- {3} and {1,2}
This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions).
If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum:
- {1,6,7} and {2,3,4,5}
- {2,5,7} and {1,3,4,6}
- {3,4,7} and {1,2,5,6}
- {1,2,4,7} and {3,5,6}
Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways.
Your program must calculate the answer, not look it up from a table.
PROGRAM NAME: subset
INPUT FORMAT
The input file contains a single line with a single integer representing N, as above.
SAMPLE INPUT (file subset.in)
7
OUTPUT FORMAT
The output file contains a single line with a single integer that tells how many same-sum partitions can be made from the set {1, 2, ..., N}. The output file should contain 0 if there are no ways to make a same-sum partition.
SAMPLE OUTPUT (file subset.out)
4
类型：DP
题意：将1~n，n个数分成两组，使其和相等，求一共有多少种分法。改变数字的顺序不会增加分法。
思路：很裸的背包。如果M=n*(n+1)/2是奇数，则没有分法。如果是偶数，背包容量为M/2，dp[k] += dp[k-i],(i=1,2,...,n)计算k的时候为避免重算，
需倒着进行。
代码
/*
ID: superbi1
TASK: subset
LANG: C
*/
#include
<
stdio.h
>
#include
<
string
.h
>
int
main()
{
unsigned
int
dp[
391
];
int
i, j, n, M;
FILE
*
in
=
fopen(
"
subset.in
"
,
"
r
"
);
FILE
*
out
=
fopen(
"
subset.out
"
,
"
w
"
);
fscanf(
in
,
"
%d
"
,
&
n);
M
=
n
*
(n
+
1
)
/
2
;
if
(M
&
1
) {
fprintf(
out
,
"
0\n
"
);
return
0
;
}
M
/=
2
;
memset(dp,
0
,
sizeof
(dp));
dp[
0
]
=
1
;
for
(i
=
1
; i
<=
n; i
++
) {
for
(j
=
M
-
i; j
>=
0
; j
--
)
if
(dp[j]) {
dp[j
+
i]
+=
dp[j];
}
}
fprintf(
out
,
"
%u\n
"
, dp[M]
/
2
);
return
0
;
}