CF364

链接：http://codeforces.com/contest/364

A:Matrix

关于矩阵。。。

Let's notice that sum in the rectangle (x1, y1, x2, y2) is sum(x1, x2)·sum(y1, y2). Where sum(l, r) = sl + sl + 1 + ... + sr. Then, we have to calc sum(l, r) for every pair (l, r) and count how many segments give us sum x for any possible x (0 ≤ x ≤ 9·|s|). In the end we should enumerate sum on segemnt [x1, x2] and find . There is corner case a = 0.

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
#define LL long long
const int maxn=4010;
LL n;
char s[maxn];
LL a[maxn][maxn],sum[maxn][maxn];
int main()
{
    cin>>n>>s+1;
    int len=strlen(s+1);
    for(int i=1;i<=len;i++)
        for(int j=1;j<=len;j++)
        a[i][j]=(s[i]-'0')*(s[j]-'0');
    for(int i=1;i<=len;i++)
        for(int j=1;j<=len;j++)
            sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+a[i][j];
            LL ans=0;
    for(int i=1;i<=len;i++)
        for(int j=1;j<=len;j++)
            for(int k=1;k<=i;k++)
                for(int l=1;l<=j;l++)
        if(sum[i][j]+sum[k-1][l-1]-sum[i][l-1]-sum[k-1][j]==n) ans++;
           cout<<ans<<endl;
}

#include <cstdio>
#include <numeric>
#include <iostream>
#include <vector>
#include <set>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <bitset>
#include <queue>
#include <sstream>
#include <deque>

using namespace std;

#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i = 0; i < (n); i++)
#define re return
#define fi first
#define se second
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define sqr(x) ((x) * (x))
#define sqrt(x) sqrt(abs(x))
#define y0 y3487465
#define y1 y8687969
#define fill(x,y) memset(x,y,sizeof(x))
                         
typedef vector<int> vi;
typedef long long ll;
typedef long double ld;
typedef double D;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef vector<vi> vvi;

template<class T> T abs(T x) { re x > 0 ? x : -x; }

const int N = 40000;

int n;
int m;
string s;

int cnt[N];

int main () {
    cin >> m >> s;
    n = sz (s);
    for (int i = 0; i < n; i++) {
        int cur = 0;
        for (int j = i; j < n; j++) {
            cur += s[j] - '0';
            cnt[cur]++;
        }
    }
    ll ans = 0;
    if (m == 0) {
        int all = n * (n + 1) / 2;
        ans += (ll)all * all - (ll)(all - cnt[0]) * (all - cnt[0]);
    } else {
        for (int i = 1; i * i <= m && i < N; i++)
            if (m % i == 0 && m / i < N) {
                int j = m / i;
                if (i != j) ans += (ll)2 * cnt[i] * cnt[j]; else ans += (ll)cnt[i] * cnt[i];
            }
    }
    cout << ans << endl;
    return 0;
}

#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <cmath>
#include <cstring>
#include <string>
#include <iostream>
#include <complex>
#include <sstream>
using namespace std;
 
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
typedef vector<int> VI;
typedef pair<int,int> PII;
 
#define REP(i,n) for(int i=0;i<(n);++i)
#define SIZE(c) ((int)((c).size()))
#define FOR(i,a,b) for (int i=(a); i<(b); ++i)
#define FOREACH(i,x) for (__typeof((x).begin()) i=(x).begin(); i!=(x).end(); ++i)
#define FORD(i,a,b) for (int i=(a)-1; i>=(b); --i)
#define ALL(v) (v).begin(), (v).end()
 
#define pb push_back
#define mp make_pair
#define st first
#define nd second

LL gcd(LL a, LL b) {
    return b ? gcd(b,a%b) : a;
}

char S[5000];
int su[5000];
int main() {
    int A;
    scanf("%d", &A);
    scanf("%s", S);
    int N = strlen(S);
    REP(i,N) S[i] -= '0';
    
    map<int,int> C;
    su[0] = 0;
    FOR(i,1,N+1) su[i] = S[i-1] + su[i-1];
    
    map<int,int> M;
    REP(i,N+1)REP(j,i) M[su[i] - su[j]]++;

    if (A == 0) {
        LL Y = (N + 1) * N / 2;
        LL X = M[0];
        cout << (Y * Y - (Y - X) * (Y - X)) <<  endl;
        return 0;
    }

    LL result = 0;
    for (int a = 1; a * a <= A; ++a) {
        if (A % a) continue;
        LL X = M[a];
        LL Y = M[A / a];
    
        result += X * Y;
        if (a * a != A) result += X * Y;
    }
    
    cout << result << endl;
}