题目背景
这是一个, 被战斗因果所支配的将宇宙的命运钻开风洞的男人的故事—— 在遥远的太古时代......, 某群螺旋族人发现了一个重大的事实。螺旋力进化的最终结果, 就是宇宙的灭亡, 等待他们的只有螺旋神怒“Spiral-Nemesis”
对此感到恐惧的这群螺旋族人, 为了防止宇宙的崩坏, 消灭了大量的持有螺旋力量的族人,并将所剩无几的生命囚禁于宇宙的角落里。同时, 他们也停止了自身的进化, 将自己封闭在了不同于次元轴世界的隔绝宇宙中。从此以 Anti-Spiral自称的他们, 继续与幸存下来的螺旋族人战斗着。众多螺旋的战士向Anti-Spiral发起挑战并败下阵来。西蒙他们将地球托付给罗修之后, 乘上被命名为“超银河大 Gurren”的流星圣殿,向Anti-Spiral的母星进发。所在场所长期不明的Anti-Spiral的母星, 由西蒙送给妮亚的戒指而回应了螺旋索敌反应。
题目描述
螺旋索敌反应会传达回一个长度为n的序列a1?an。
西蒙知道Anti-Spiral的母星所处在的宇宙的维度恰好就是这n个数两两配对所能形成的最多的质数的个数。
所谓配对是指选出一个和一个进行配对,一个配对将形成一个新的数ai+aj。(注:ai+aj可以重复)
对于每一个ai最多出现在一个配对之中,当然也可以不出现在任意一个配对中。
现在西蒙想快速知道Anti-Spiral的母星所处在的宇宙的维度
输入输出格式
输入格式:
第一行输入一个数,n
第二行n个数,a1...an
输出格式:
一行一个数,表示 Anti-Spiral 的母星所处在的宇宙的维度
输入输出样例
输入样例#1:
5 2 9 11 12 37
输出样例#1:
2
说明
2和9配对 11和12配对
【数据规模及约定】 对于 30% 的数据,n ≤ 10,2 ≤ai ≤ 100。 对于另外 30% 的数据,n ≤ 40,2 ≤ai ≤ 200。 对于 100% 的数据,n ≤ 40,2 ≤ ai≤ 1000。
这道题可以看出只有一奇一偶两个数可以凑成一个质数。所以可以把这个分成奇数和偶数两个集合,如果可以配对,连一下边。最后找最大匹配即可。
代码如下:
1 #include<cstdio> 2 #include<cmath> 3 #include<algorithm> 4 #include<queue> 5 #include<cstring> 6 using namespace std; 7 typedef long long ll; 8 const ll maxn = 1e5 + 7; 9 bool prime[2005] = {0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 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1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0}; 10 ll read() 11 { 12 ll a = 0,b = 1; 13 char c = getchar(); 14 while(c < ‘0‘ or c > ‘9‘) 15 { 16 if(c == ‘-‘) b = -1; 17 c = getchar(); 18 } 19 while(c >= ‘0‘ and c <= ‘9‘) 20 { 21 a = a * 10 + c - ‘0‘; 22 c = getchar(); 23 } 24 return a * b; 25 } 26 struct node 27 { 28 int c,v,nxt; 29 }e[maxn]; 30 ll head[maxn],cnt = 0,ans,n,pri[maxn],vis[maxn],nxt[maxn],a[maxn],b[maxn],numa,numb,mp[45][45]; 31 32 bool find(ll x) 33 { 34 for(int i=1; i<=numb; i++) 35 { 36 if(vis[i] == 0 and mp[x][i]) 37 { 38 vis[i] = 1; 39 if(pri[i] == 0 || find(pri[i])) 40 { 41 pri[i] = x; 42 return true; 43 } 44 } 45 } 46 return false; 47 } 48 int main() 49 { 50 n = read(); 51 for(int i=1; i<=n; i++) 52 { 53 ll t = read(); 54 if(t % 2 == 1) 55 a[++numa] = t; 56 else 57 b[++numb] = t; 58 } 59 for(int i=1; i<=numa; i++) 60 { 61 for(int j=1; j<=numb; j++) 62 { 63 if(prime[a[i] + b[j]] == 1) 64 { 65 mp[i][j] = 1; 66 } 67 } 68 } 69 for(int i=1; i<=numa; i++) 70 { 71 memset(vis,0,sizeof(vis)); 72 if(find(i)) 73 ans++; 74 } 75 printf("%lld",ans); 76 }
原文地址:https://www.cnblogs.com/qmcp/p/9579891.html