这道题比较水。大概思路是使用循环，之后检查 \(a_{i}\) 是否达到 \(x\) 且 \(b_{i}\) 是否达到 \(y\) 且 \(a_{i} + b_{i}\) 是否达到 \(z\)。
代码如下。
#include <iostream>
namespace io{
template <typename T> inline void read(T& x){x = 0; bool f = false; char ch = getchar(); while(ch < '0' || ch > '9'){if(ch == '-') f = !f; ch = getchar();}while(ch >= '0' && ch <= '9'){x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();} x = (f ? -x : x); return;}
template <typename T, typename... Args> inline void read(T& x, Args&...x_){read(x), read(x_...);}
template <typename T> inline void put(T x){if(x < 0) putchar('-'), x = -x; if(x > 9) put(x / 10); putchar(x % 10 + '0'); return ;}
template <typename T> inline void write(T x){put(x);}
template <typename T, typename... Args> inline void write(T x, Args...x_){write(x), write(x_...);}
inline void newb(){putchar(' ');}
}
namespace code{
int n, x, y, z, a[101], b[101], cnt = 0;
int main(){
io::read(n, x, y, z);
for(int i = 1; i <= n; i++) io::read(a[i], b[i]);
for(int i = 1; i <= n; i++) if(a[i] >= x && b[i] >= y && a[i] + b[i] >= z) cnt++;
io::write(cnt);
return 0;
}
}
int main(){
code::main();
return 0;
}