中国剩余定理，又名孙子定理o(*≧▽≦)ツ
能求解什么问题呢？
问题：
一堆物品
3个3个分剩2个
5个5个分剩3个
7个7个分剩2个
问这个物品有多少个
解这题，我们需要构造一个答案
我们需要构造这个答案
5*7*inv(5*7, 3) % 3 = 1
3*7*inv(3*7, 5) % 5 = 1
3*5*inv(3*5, 7) % 7 = 1
这3个式子对不对，别告诉我逆元你忘了(*′?｀*)，忘了的人请翻阅前几章复习
然后两边同乘你需要的数
2 * 5*7*inv(5*7, 3) % 3 = 2
3 * 3*7*inv(3*7, 5) % 5 = 3
2 * 3*5*inv(3*5, 7) % 7 = 2
令
a = 2 * 5*7*inv(5*7, 3)
b = 3 * 3*7*inv(3*7, 5)
c = 2 * 3*5*inv(3*5, 7)
那么
a % 3 = 2
b % 5 = 3
c % 7 = 2
其实答案就是a+b+c
因为
a%5 = a%7 = 0 因为a是5的倍数，也是7的倍数
b%3 = b%7 = 0 因为b是3的倍数，也是7的倍数
c%3 = c%5 = 0 因为c是3的倍数，也是5的倍数
所以
(a + b + c) % 3 = (a % 3) + (b % 3) + (c % 3) = 2 + 0 + 0 = 2
(a + b + c) % 5 = (a % 5) + (b % 5) + (c % 5) = 0 + 3 + 0 = 3
(a + b + c) % 7 = (a % 7) + (b % 7) + (c % 7) = 0 + 0 + 2 = 2
你看你看，答案是不是a+b+c(??ω?)??，完全满足题意
但是答案，不只一个，有无穷个，每105个就是一个答案（105 = 3 * 5 * 7）
根据计算，答案等于233,233%105 = 23
如果题目问你最小的那个答案，那就是23了
以下抄自百度百科
1 //n个方程：x=a[i](mod m[i]) (0<=i<n)
2 LL china(int n, LL *a, LL *m){
3 LL M = 1, ret = 0;
4 for(int i = 0; i < n; i ++) M *= m[i];
5 for(int i = 0; i < n; i ++){
6 LL w = M / m[i];
7 ret = (ret + w * inv(w, m[i]) * a[i]) % M;
8 }
9 return (ret + M) % M;
10 }
要不要来一道题试试手？
poj 1006
http://poj.org/problem?id=1006
问题描述：
人自出生起就有体力，情感和智力三个生理周期，分别为23，28和33天。一个周期内有一天为峰值，在这一天，人在对应的方面（体力，情感或智力）表现最好。通常这三个周期的峰值不会是同一天。现在给出三个日期，分别对应于体力，情感，智力出现峰值的日期。然后再给出一个起始日期，要求从这一天开始，算出最少再过多少天后三个峰值同时出现。
分析：
因为23 = 23
28 = 2*2*7
33 = 3*11
满足两两互质关系，所以直接套模板就好了
AC代码：
#include<cstdio>
typedef long long LL;
const int N = 100000 + 5;
void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){
if (!b) {d = a, x = 1, y = 0;}
else{
ex_gcd(b, a % b, y, x, d);
y -= x * (a / b);
}
}
LL inv(LL t, LL p){//如果不存在，返回-1
LL d, x, y;
ex_gcd(t, p, x, y, d);
return d == 1 ? (x % p + p) % p : -1;
}
LL china(int n, LL *a, LL *m){//中国剩余定理
LL M = 1, ret = 0;
for(int i = 0; i < n; i ++) M *= m[i];
for(int i = 0; i < n; i ++){
LL w = M / m[i];
ret = (ret + w * inv(w, m[i]) * a[i]) % M;
}
return (ret + M) % M;
}
int main(){
LL p[3], r[3], d, ans, MOD = 21252;
int cas = 0;
p[0] = 23; p[1] = 28; p[2] = 33;
while(~scanf("%I64d%I64d%I64d%I64d", &r[0], &r[1], &r[2], &d) && (~r[0] || ~r[1] || ~r[2] || ~d)){
ans = ((china(3, r, p) - d) % MOD + MOD) % MOD;
printf("Case %d: the next triple peak occurs in %I64d days.\n", ++cas, ans ? ans : 21252);
}
}
1 #include<cstdio>
2 typedef long long LL;
3 const int N = 100000 + 5;
4 void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){
5 if (!b) {d = a, x = 1, y = 0;}
6 else{
7 ex_gcd(b, a % b, y, x, d);
8 y -= x * (a / b);
9 }
10 }
11 LL inv(LL t, LL p){//如果不存在，返回-1
12 LL d, x, y;
13 ex_gcd(t, p, x, y, d);
14 return d == 1 ? (x % p + p) % p : -1;
15 }
16 LL china(int n, LL *a, LL *m){//中国剩余定理
17 LL M = 1, ret = 0;
18 for(int i = 0; i < n; i ++) M *= m[i];
19 for(int i = 0; i < n; i ++){
20 LL w = M / m[i];
21 ret = (ret + w * inv(w, m[i]) * a[i]) % M;
22 }
23 return (ret + M) % M;
24 }
25 int main(){
26 LL p[3], r[3], d, ans, MOD = 21252;
27 int cas = 0;
28 p[0] = 23; p[1] = 28; p[2] = 33;
29 while(~scanf("%I64d%I64d%I64d%I64d", &r[0], &r[1], &r[2], &d) && (~r[0] || ~r[1] || ~r[2] || ~d)){
30 ans = ((china(3, r, p) - d) % MOD + MOD) % MOD;
31 printf("Case %d: the next triple peak occurs in %I64d days.\n", ++cas, ans ? ans : 21252);
32 }
33
34 }
当然，这个中国剩余定理只是基础，面对更强大的敌人，我们要有更强的武器
比如，m1,m2, ... ,mn两两不保证互质，辣怎么办(っ °Д °)っ
别怕，看我接着抛代码
1 #include<cstdio>
2 #include<algorithm>
3 using namespace std;
4 typedef long long LL;
5 typedef pair<LL, LL> PLL;
6 PLL linear(LL A[], LL B[], LL M[], int n) {//求解A[i]x = B[i] (mod M[i]),总共n个线性方程组
7 LL x = 0, m = 1;
8 for(int i = 0; i < n; i ++) {
9 LL a = A[i] * m, b = B[i] - A[i]*x, d = gcd(M[i], a);
10 if(b % d != 0) return PLL(0, -1);//答案不存在，返回-1
11 LL t = b/d * inv(a/d, M[i]/d)%(M[i]/d);
12 x = x + m*t;
13 m *= M[i]/d;
14 }
15 x = (x % m + m ) % m;
16 return PLL(x, m);//返回的x就是答案，m是最后的lcm值
17 }
这个代码我不给予解释（因为我不会，哇哈哈哈╰(*°▽°*)╯）
遇到需要的题就去套模板吧
（想知道代码原理的去百度吧，或者看《挑战程序设计竞赛》，我模板是从书里抄来经过杰哥修改的）
比如poj 2891
http://poj.org/problem?id=2891
【题目大意】
给出k个模方程组：x mod ai = ri。求x的最小正值。如果不存在这样的x，那么输出-1.
【题目分析】
由于这道题目里面的ai、ri之间不满足两两互质的性质，所以不能用中国剩余定理直接求解。
辣么。。。。愉快的套这个模板吧
AC代码如下:
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef pair<LL, LL> PLL;
LL a[100000], b[100000], m[100000];
LL gcd(LL a, LL b){
return b ? gcd(b, a%b) : a;
}
void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){
if (!b) {d = a, x = 1, y = 0;}
else{
ex_gcd(b, a % b, y, x, d);
y -= x * (a / b);
}
}
LL inv(LL t, LL p){//如果不存在，返回-1
LL d, x, y;
ex_gcd(t, p, x, y, d);
return d == 1 ? (x % p + p) % p : -1;
}
PLL linear(LL A[], LL B[], LL M[], int n) {//求解A[i]x = B[i] (mod M[i]),总共n个线性方程组
LL x = 0, m = 1;
for(int i = 0; i < n; i ++) {
LL a = A[i] * m, b = B[i] - A[i]*x, d = gcd(M[i], a);
if(b % d != 0) return PLL(0, -1);//答案，不存在，返回-1
LL t = b/d * inv(a/d, M[i]/d)%(M[i]/d);
x = x + m*t;
m *= M[i]/d;
}
x = (x % m + m ) % m;
return PLL(x, m);//返回的x就是答案，m是最后的lcm值
}
int main(){
int n;
while(scanf("%d", &n) != EOF){
for(int i = 0; i < n; i ++){
a[i] = 1;
scanf("%d%d", &m[i], &b[i]);
}
PLL ans = linear(a, b, m, n);
if(ans.second == -1) printf("-1\n");
else printf("%I64d\n", ans.first);
}
}
1 #include<cstdio>
2 #include<algorithm>
3 using namespace std;
4 typedef long long LL;
5 typedef pair<LL, LL> PLL;
6 LL a[100000], b[100000], m[100000];
7 LL gcd(LL a, LL b){
8 return b ? gcd(b, a%b) : a;
9 }
10 void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){
11 if (!b) {d = a, x = 1, y = 0;}
12 else{
13 ex_gcd(b, a % b, y, x, d);
14 y -= x * (a / b);
15 }
16 }
17 LL inv(LL t, LL p){//如果不存在，返回-1
18 LL d, x, y;
19 ex_gcd(t, p, x, y, d);
20 return d == 1 ? (x % p + p) % p : -1;
21 }
22 PLL linear(LL A[], LL B[], LL M[], int n) {//求解A[i]x = B[i] (mod M[i]),总共n个线性方程组
23 LL x = 0, m = 1;
24 for(int i = 0; i < n; i ++) {
25 LL a = A[i] * m, b = B[i] - A[i]*x, d = gcd(M[i], a);
26 if(b % d != 0) return PLL(0, -1);//答案，不存在，返回-1
27 LL t = b/d * inv(a/d, M[i]/d)%(M[i]/d);
28 x = x + m*t;
29 m *= M[i]/d;
30 }
31 x = (x % m + m ) % m;
32 return PLL(x, m);//返回的x就是答案，m是最后的lcm值
33 }
34 int main(){
35 int n;
36 while(scanf("%d", &n) != EOF){
37 for(int i = 0; i < n; i ++){
38 a[i] = 1;
39 scanf("%d%d", &m[i], &b[i]);
40 }
41 PLL ans = linear(a, b, m, n);
42 if(ans.second == -1) printf("-1\n");
43 else printf("%I64d\n", ans.first);
44 }
45 }
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef pair<LL, LL> PLL;
LL a[100000], b[100000], m[100000];
LL gcd(LL a, LL b){
return b ? gcd(b, a%b) : a;
}
void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){
if (!b) {d = a, x = 1, y = 0;}
else{
ex_gcd(b, a % b, y, x, d);
y -= x * (a / b);
}
}
LL inv(LL t, LL p){//如果不存在，返回-1
LL d, x, y;
ex_gcd(t, p, x, y, d);
return d == 1 ? (x % p + p) % p : -1;
}
PLL linear(LL A[], LL B[], LL M[], int n) {//求解A[i]x = B[i] (mod M[i]),总共n个线性方程组
LL x = 0, m = 1;
for(int i = 0; i < n; i ++) {
LL a = A[i] * m, b = B[i] - A[i]*x, d = gcd(M[i], a);
if(b % d != 0) return PLL(0, -1);//答案，不存在，返回-1
LL t = b/d * inv(a/d, M[i]/d)%(M[i]/d);
x = x + m*t;
m *= M[i]/d;
}
x = (x % m + m ) % m;
return PLL(x, m);//返回的x就是答案，m是最后的lcm值
}
int main(){
int n;
while(scanf("%d", &n) != EOF){
for(int i = 0; i < n; i ++){
a[i] = 1;
scanf("%d%d", &m[i], &b[i]);
}
PLL ans = linear(a, b, m, n);
if(ans.second == -1) printf("-1\n");
else printf("%I64d\n", ans.first);
}
}
