UVA 12169 Disgruntled Judge (扩展欧几里得)
2016-10-26 23:54
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大体题意:
给你n 个数 x1,x3,,,x2n-1
让你确定出x2,x4,,,x2n来!
x数列每一项满足 xi = ((x(i-1) * a ) + b ) % 10001;
思路:
我们枚举a,因为对10001取模,所以a肯定在0~10000以内有解!
那么我们就枚举a,对于每一项a,我们根据x3和x1求出b来,然后开始枚举每一项看看这个a和b 是否合适!合适就输出终止枚举!
判断合适的方法是 跑一遍递推式,当发现与输入冲突时,不合适! 都不冲突 则合适!
这个题难点就在于如何求出b来!
这考察了 扩展欧几里得:
根据定义:
将x2代入x3得:
移向得:
形如Ax + By = C的形式!
当C是gcd(A,B)的倍数时有解!
详细见代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll mod = 10001;
ll x[mod],bb,n;
void gcd(ll a,ll b,ll& d,ll& x,ll& y){
if (!b) {d = a; x=1; y = 0;}
else { gcd(b,a%b,d,y,x); y-= x*(a/b); }
}
bool judge(ll a){
ll A = mod;
ll B = -(a+1);
ll C = a*a*x[1] - x[3];
ll g,xx,y;
gcd(A,B,g,xx,y);
if (C % g) return false;
bb = y*C/g;
for (int i = 2; i <= 2*n; ++i){
if (i & 1){
if (x[i] != (a*x[i-1]+bb)%mod) return false;
}
else {
x[i] = (a*x[i-1]+bb)%mod;
}
}
return true;
}
int main(){
while(~scanf("%lld",&n)){
for (int i = 1; i <= 2*n; i += 2) scanf("%lld",x+i);
for (ll a = 0; a < mod; ++a){
if (judge(a)){
for (int i = 2; i <= 2*n; i += 2){
printf("%lld\n",x[i]%mod);
}
break;
}
}
}
return 0;
}
给你n 个数 x1,x3,,,x2n-1
让你确定出x2,x4,,,x2n来!
x数列每一项满足 xi = ((x(i-1) * a ) + b ) % 10001;
思路:
我们枚举a,因为对10001取模,所以a肯定在0~10000以内有解!
那么我们就枚举a,对于每一项a,我们根据x3和x1求出b来,然后开始枚举每一项看看这个a和b 是否合适!合适就输出终止枚举!
判断合适的方法是 跑一遍递推式,当发现与输入冲突时,不合适! 都不冲突 则合适!
这个题难点就在于如何求出b来!
这考察了 扩展欧几里得:
根据定义:
将x2代入x3得:
移向得:
形如Ax + By = C的形式!
当C是gcd(A,B)的倍数时有解!
详细见代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll mod = 10001;
ll x[mod],bb,n;
void gcd(ll a,ll b,ll& d,ll& x,ll& y){
if (!b) {d = a; x=1; y = 0;}
else { gcd(b,a%b,d,y,x); y-= x*(a/b); }
}
bool judge(ll a){
ll A = mod;
ll B = -(a+1);
ll C = a*a*x[1] - x[3];
ll g,xx,y;
gcd(A,B,g,xx,y);
if (C % g) return false;
bb = y*C/g;
for (int i = 2; i <= 2*n; ++i){
if (i & 1){
if (x[i] != (a*x[i-1]+bb)%mod) return false;
}
else {
x[i] = (a*x[i-1]+bb)%mod;
}
}
return true;
}
int main(){
while(~scanf("%lld",&n)){
for (int i = 1; i <= 2*n; i += 2) scanf("%lld",x+i);
for (ll a = 0; a < mod; ++a){
if (judge(a)){
for (int i = 2; i <= 2*n; i += 2){
printf("%lld\n",x[i]%mod);
}
break;
}
}
}
return 0;
}
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