假设S是一个具有模数为n的t个元素的集合。确实有2 ^ t个任意长度的子集。说明PARI / GP程序,其找到不同元素的最小子集U(就长度而言),使得U中所有元素的总和为0模n。编写一个通过强力搜索的程序很容易,但是当t和n变大时,暴力是不可行的,所以希望能帮助编写一个不使用强力来解决subset sum problem这个实例的程序。
答案 0 :(得分:1)
动态方法:
def isSubsetSum(st, n, sm) :
# The value of subset[i][j] will be
# true if there is a subset of
# set[0..j-1] with sum equal to i
subset=[[True] * (sm+1)] * (n+1)
# If sum is 0, then answer is true
for i in range(0, n+1) :
subset[i][0] = True
# If sum is not 0 and set is empty,
# then answer is false
for i in range(1, sm + 1) :
subset[0][i] = False
# Fill the subset table in botton
# up manner
for i in range(1, n+1) :
for j in range(1, sm+1) :
if(j < st[i-1]) :
subset[i][j] = subset[i-1][j]
if (j >= st[i-1]) :
subset[i][j] = subset[i-1][j] or subset[i - 1][j-st[i-1]]
"""uncomment this code to print table
for i in range(0,n+1) :
for j in range(0,sm+1) :
print(subset[i][j],end="")
print(" ")"""
return subset[n][sm];
答案 1 :(得分:0)
我从here获得了此代码。我不知道天气似乎有效。
function getSummingItems(a,t){
return a.reduce((h,n) => Object.keys(h)
.reduceRight((m,k) => +k+n <= t ? (m[+k+n] = m[+k+n] ? m[+k+n].concat(m[k].map(sa => sa.concat(n)))
: m[k].map(sa => sa.concat(n)),m)
: m, h), {0:[[]]})[t];
}
var arr = Array(20).fill().map((_,i) => i+1), // [1,2,..,20]
tgt = 42,
res = [];
console.time("test");
res = getSummingItems(arr,tgt);
console.timeEnd("test");
console.log("found",res.length,"subsequences summing to",tgt);
console.log(JSON.stringify(res));
&#13;