X問題
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5621 Accepted Submission(s): 1919
做了幾道發現擴展歐幾里得的簡單題貌似就是這個套路了- -
要注意的就是判斷,是否滿足a*x+b*y=c中的c是gcd(a, b)的倍數,不滿足則輸出0,最後還要注意一下最小整數解是否大於給定的N,以及是否爲0,題目要求正整數,捨去0即可
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<functional>
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CL(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long LL;
const int maxn = 1e5+10;
const int MOD = 1e9+7;
int n[20], m[20];
void ex_gcd(int a, int b, int& d, int& x, int& y){
if(!b){
d = a;
x = 1;
y = 0;
}
else{
ex_gcd(b, a%b, d, y, x);
y -= x*(a/b);
}
}
int main(){
int T, d, x, y, m1, m2, m3, n1, n2, n3, v, N, M;
int kcase = 1;
scanf("%d", &T);
while(T--){
scanf("%d%d", &N, &M);
for(int i = 1; i <= M; i++) scanf("%d", &m[i]);
for(int i = 1; i <= M; i++) scanf("%d", &n[i]);
m1 = m[1]; n1 = n[1];
bool flag = false;
for(int i = 2; i <= M; i++){
m2 = m[i]; n2 = n[i];
ex_gcd(m1, m2, d, x, y);
LL c, t, k;
c = n2-n1;
if(c % d){
flag = true;
break;
}
c /= d;
t = m2/d;
x *= c;
k = (x%t+t)%t;
n1 += m1*k;
m1 = m1/d*m2;
n1 = (n1%m1+m1)%m1;
}
if(flag || n1 > N) printf("0\n");
else{
if(!n1) printf("%d\n", N/m1);
else printf("%d\n", (N-n1)/m1 + 1);
}
}
return 0;
}