题意：http://codeforces.com/contest/1323/problem/B
思路：
先预处理可能的矩阵，用pair配对配好
假设竖着的有n根，横着的有m根，一共n*m个矩形。也就是说每两根都有机会两两结合
那就把单独一根的数量算好，最后相乘就行了
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr strcat 13 #include <string> 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 #include <cassert> 21 #include <iomanip> 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 25 //****************** 26 clock_t __START,__END; 27 double __TOTALTIME; 28 void _MS(){__START=clock();} 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;} 30 //*********************** 31 #define rint register int 32 #define fo(a,b,c) for(rint a=b;a<=c;++a) 33 #define fr(a,b,c) for(rint a=b;a>=c;--a) 34 #define mem(a,b) memset(a,b,sizeof(a)) 35 #define pr printf 36 #define sc scanf 37 #define ls rt<<1 38 #define rs rt<<1|1 39 typedef pair<int,int> PII; 40 typedef vector<int> VI; 41 typedef unsigned long long ull; 42 typedef long long ll; 43 typedef double db; 44 const db E=2.718281828; 45 const db PI=acos(-1.0); 46 const ll INF=(1LL<<60); 47 const int inf=(1<<30); 48 const db ESP=1e-9; 49 const int mod=(int)1e9+7; 50 const int N=(int)1e6+10; 51 52 int cnt; 53 pair<ll,ll>pri[N]; 54 ll a[N],b[N]; 55 56 void go(int x) 57 { 58 for(int i=1;i<=cnt;++i) 59 { 60 if(pri[i].first<=x) 61 a[i]+=x+1-pri[i].first; 62 } 63 } 64 void goo(int x) 65 { 66 for(int i=1;i<=cnt;++i) 67 { 68 if(pri[i].second<=x) 69 b[i]+=x+1-pri[i].second; 70 } 71 } 72 73 int main() 74 { 75 ll n,m,k; 76 cin>>n>>m>>k; 77 for(ll i=1;i*i<=k;++i) 78 { 79 if(k%i==0) 80 { 81 pri[++cnt]={i,k/i}; 82 if(k/i!=i) 83 pri[++cnt]={k/i,i}; 84 } 85 } 86 int cc=0; 87 for(int i=1;i<=n;++i) 88 { 89 int t; 90 sc("%d",&t); 91 if(t) 92 cc++; 93 else 94 { 95 go(cc); 96 cc=0; 97 } 98 } 99 if(cc) 100 go(cc); 101 cc=0; 102 for(int i=1;i<=m;++i) 103 { 104 int t; 105 sc("%d",&t); 106 if(t) 107 cc++; 108 else 109 { 110 goo(cc); 111 cc=0; 112 } 113 } 114 if(cc) 115 goo(cc); 116 cc=0; 117 ll ans=0; 118 for(int i=1;i<=cnt;++i) 119 ans+=a[i]*b[i]; 120 pr("%lld ",ans); 121 return 0; 122 } 123 124 /**************************************************************************************/