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vjudge 上题目链接：uva 11728
其实是个数论水题，直接打表就行：
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 5 int ans[1003]; 6 inline void init(int n = 1000) { 7 for(int i = 1; i <= n; ++i) { 8 int sum = 0; 9 for(int j = 1; j <= i; ++j) 10 if(i % j == 0) sum += j; 11 if(sum <= 1000) ans[sum] = i; 12 } 13 } 14 15 int main() { 16 int s,Case = 0; 17 init(); 18 while(~scanf("%d",&s),s) 19 printf("Case %d: %d\n",++Case, !ans[s]? -1: ans[s]); 20 return 0; 21 }
可我一开始却杀鸡用了牛刀，想得超复杂，就因为大白书上提示"唯一分解定理，回溯"什么的，于是我就按着这样的思路去做了，结果写得超复杂，感觉过了也没有任何意义了。。。
1 #include<cstdio> 2 #include<cstring> 3 #include<vector> 4 #include<algorithm> 5 using namespace std; 6 7 bool vis[1060]; 8 int pri[200]; 9 inline int init_pri(int n = 1050) { 10 int c = 0; 11 for(int i = 2; i <= n; ++i) 12 if(!vis[i]) { 13 pri[++c] = i; 14 for(int j = i << 1; j <= n; j += i) vis[j] = 1; 15 } 16 return c; 17 } 18 19 struct node { 20 int sum, p, r, pow; 21 node() {} 22 node(int sum, int p, int r, int pow): sum(sum), p(p), r(r), pow(pow) {} 23 bool operator < (const node & n2) const { 24 if(sum == n2.sum) return p < n2.p; 25 return sum < n2.sum; 26 } 27 void print() const { 28 printf("sum = %d p = %d r = %d pow = %d\n",sum,p,r,pow); 29 } 30 }; 31 32 vector<node> fac; 33 34 inline void init() { 35 init_pri(); 36 for(int i = 1; pri[i] <= 1020; ++i) { 37 int sum = 1, mul = 1, len = 0; 38 while(sum <= 1020) { 39 mul *= pri[i]; 40 sum += mul; 41 if(sum > 1020) break; 42 fac.push_back(node(sum, pri[i], ++len, mul)); 43 } 44 } 45 sort(fac.begin(), fac.end()); 46 } 47 48 int find(const vector<node> &c, int low, int up, int x) { 49 int mid; 50 while(low <= up) { 51 mid = low + up >> 1; 52 if(c[mid].sum == x) return mid; 53 else if(x < c[mid].sum) up = mid - 1; 54 else low = mid + 1; 55 } 56 return low; 57 } 58 59 int ans; 60 bool sign[1030]; 61 62 void dfs(int s, int id, int n) { 63 if(s == 1) { 64 if(n > ans) ans = n; 65 return ; 66 } 67 if(s <= 0 || id < 0) return ; 68 if(!sign[fac[id].p] && s % fac[id].sum == 0) { 69 sign[fac[id].p] = 1; 70 dfs(s / fac[id].sum, id - 1, n * fac[id].pow); 71 sign[fac[id].p] = 0; 72 } 73 dfs(s, id - 1, n); 74 } 75 76 int main() { 77 int s, Case = 0; 78 init(); 79 while(~scanf("%d",&s), s) { 80 printf("Case %d: ",++Case); 81 if(s == 1) { 82 puts("1"); 83 continue; 84 } 85 if(s == 2) { 86 puts("-1"); 87 continue; 88 } 89 ans = -1; 90 int id = find(fac, 0, fac.size() - 1, s); 91 if(fac[id].sum > s) --id; 92 memset(sign, 0, sizeof(sign)); 93 dfs(s, id, 1); 94 printf("%d\n",ans); 95 } 96 return 0; 97 }
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原文地址：http://www.cnblogs.com/Newdawn/p/4712842.html
