題目大意：

　給你一個數A，以及一串全是數字的字串以構造矩陣C，C[i][j]=a[i]*a[j](a[k]表示字串中第k位所代表的數字).請你求出權值之和恰好為A的子矩陣個數.

solution：

　此題比較有意思.題目要我們求的答案即滿足$(\sum_{i=x}^{u}\sum_{j=y}^{v} C[i][j]) ==A$的四元組[x,y,u,v]個數.接下來我們分析一下這個式子：$$\sum_{i=x}^{u}\sum_{j=y}^{v}C[i][j]=\sum_{i=x}^{u}\sum_{j=y}^{v} a[i] \cdot a[j]=\sum_{i=1}^{u}(a[i] \cdot \sum_{j=y}^{v}a[j])=\sum_{i=x}^{u}a[i] \cdot \sum_{j=y}^{v}a[j]$$$$=(sum[u]-sum[x-1]) \cdot (sum[v]-sum[y-1])$$

code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#define R register
#define next exnttttttttttttttttttttt
#define
 
 debug puts("mlg")
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
inline ll read();
inline void write(ll x);
inline void writesp(ll x);
inline void writeln(ll x);
ll A;
char wn;
ll num[6000],n;
ll used[150000],sum[150000];
ll ans;
int main(){
    A=read();
    wn
 
=getchar();
    while(wn<'0'||wn>'9') wn=getchar();
    while(wn>='0'&&wn<='9'){
        num[++n]=wn-'0';wn=getchar();
    } 
    for(R ll i=1;i<=n;i++){
        sum[i]=sum[i-1]+num[i];
    }
    for(R ll i=1;i<=n;i++){
        for(R ll j=1;j<=i;j++){
            ++used[sum[i]-sum[j-1]];
        }
    }
    if(A==0){
        for(R ll i=0;i<=120000;i++) ans+=used[0]*used[i];
        ans<<=1;ans-=used[0]*used[0];
        writeln(ans);return 0;
    }
    for(R ll i=1;i<=120000;i++){
        if(A%i==0&&used[i]&&i*100000>=A){
            ans+=used[i]*used[A/i];
        }
    }
    writeln(ans);
}
inline ll read(){
    ll x=0,t=1;char ch=getchar();
    while(ch<'0'||ch>'9'){
        if(ch=='-') t=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*t;
}
inline void write(ll x){
    if(x<0){putchar('-');x=-x;}
    if(x<=9){putchar(x+'0');return;}
    write(x/10);putchar(x%10+'0');
}
inline void writesp(ll x){
    write(x);putchar(' ');
}
inline void writeln(ll x){
    write(x);putchar('\n');
}