codeforces A. Nastya and Rice

题目

题意：

对于每一个样例,是否满足$\exists n*(\{a-b\leq x\leq a+b\}) = (\{c-d\leq y\leq c+d\})$∃n∗({a−b≤x≤a+b})=({c−d≤y≤c+d})

思路：

我们把$n*(\{a-b\leq x\leq a+b\})$n∗({a−b≤x≤a+b})算出来，然后和$y$y进行对比，看看是否相加，不相交的情况是$n * max(x) < min(y)$n∗max(x)<min(y)和$n*min(x) > max(y)$n∗min(x)>max(y)，排除这两种就是其他的都是$Yes了$Yes了。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <deque>
#include <stack>
using namespace std;
typedef long long ll;
typedef vector<int> veci;
typedef vector<ll> vecl;
typedef pair<int, int> pii;
template <class T>
inline void read(T &ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) return ;
    while (c != '-' && (c < '0' || c > '9')) c = getchar();
    sgn = (c == '-') ? -1:1;
    ret = (c == '-') ? 0:(c - '0');
    while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return ;
}
inline void out(int x) {
    if (x > 9) out(x / 10);
    putchar(x % 10 + '0');
}
int main() {
    int t;
    read(t);
    while (t--) {
        int n, a, b, c, d;
        read(n), read(a), read(b), read(c), read(d);
        int x = (a - b) * n;
        int y = (a + b) * n;
        int xx = (c - d);
        int yy = c + d;
        if (x > yy) {
            printf("No\n");
            continue;
        } else if (xx > y) {
            printf("No\n");
            continue;
        }
        printf("Yes\n");
    }
    return 0;
}