#include"stdio.h"
#include"iostream"
#include"math.h"
#include"algorithm"

using namespace std;
typedef long long ll;
/*
void gcd(int a,int b,int &d,int &x,int &y)
{
if(b==0){d=a;x=1;y=0;return;}
int x1,y1;
gcd(b,a%b,d,x1,y1);
int t1,t2;
t1=x1;t2=y1;
x=t2;y=t1-(a/b)*t2;
}
*/
void gcd(ll a,ll b,ll &d,ll &x,ll &y)
{
if(!b){d=a;y=0;x=1;}
else{
gcd(b,a%b,d,y,x);y-=(a/b)*x;
}
}
int main()
{
ll m,n,l,x,y;
while(scanf("%I64d%I64d%I64d%I64d%I64d",&x,&y,&m,&n,&l)==5)
{
ll a,b,c;
a=m-n;b=l;c=(y-x);
if(a<0)
{
a=-a;c=-c;
}
ll sx,sy,g;
gcd(a,b,g,sx,sy);
if(c%g)
{
puts("Impossible");continue;
}
b/=g;
c/=g; //倍数

ll t=c*sx;
printf("%d\n",(t+b)%b);

}
return 0;
}

测试数据 #0: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 560 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 556 KiB, score = 10
Accepted, time = 0 ms, mem = 560 KiB, score = 100

测试数据 #0: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 548 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 552 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 552 KiB, score = 10
Accepted, time = 15 ms, mem = 552 KiB, score = 100

http://www.cnblogs.com/Coolxxx/p/5348092.html

原题目等价于求方程 ax=b(mod l)

我的步骤如下：
一、化简系数，三个系数a,b,l同时除以它们的最大公约数。
二、用辗转相除法，计算出am+ln=gcd(a,l)中的m和n。
三、由二中的m，有
am = gcd(a,l) (mod l)
=> am*(b/gcd(a,l)) = gcd(a,l)*(b/gcd(a,l)) (mod l)
=> a*(mb/gcd(a,l)) = b (mod l)
=> x = (mb / gcd(a,l)) (mod l)

http://www.cnblogs.com/oierforever/p/4764152.html

#include <vector>
#include <list>
#include <limits.h>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <string.h>
#include <stdlib.h>
using namespace std;

long long L;

long long extended_euclidean(long long a, long long b, long long &x, long long &y){
if(b==0){
x = 1;
y = 0;
return a;
}
long long g = extended_euclidean(b, a%b, x, y);
long long tmp = x;
x = y;
y = tmp - (a/b)*y;
return g;
}

int main(){

long long x, y, m, n;
cin>>x>>y>>m>>n>>L;
long long a = (m-n+L)%L;
long long b = L;
long long c = (y-x+L)%L;
long long g = extended_euclidean(a, b, x, y);
if(c%g != 0){
cout<<"Impossible"<<endl;
return 0;
}
long long k = c/g;
x = (x*k);
if(x < 0)
x = (L/g-(abs(x)%(L/g)));
if((x-L/g) > 0)
x = x%(L/g);
cout<<x<<endl;
return 0;
}
一句话：**扩欧**
头文件是模板，程序是复制的，搬运了一下
原网页

#include<iostream>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

void gcd(int a,int b,int &d,int&x,int &y){
if (b == 0){
d = a;
x = 1;
y = 0;
}else{
gcd(b, a%b, d, y, x);
y -= x*(a / b);
}
}

int main(){
freopen("in.txt", "r", stdin);
int x1, x2, v1, v2,len;
cin >> x1 >> x2 >> v1 >> v2 >> len;
int dx, dv;
v1 %= len;
v2 %= len;
x1 %= len;
x2 %= len;
if (v1 < v2){
swap(v1, v2);
swap(x1, x2);
}
dv = v1 - v2;
if (x1<x2){
dx = x2 - x1;
}
else{
dx = x2+len-x1;
}
int d, k,p;
gcd(dv,len, d,k,p);
if (dx%d){
cout << "Impossible" << endl;
}
else{
dv /= d;
len /= d;
if (k > 0){

cout << (dx*k / d)%len << endl;
}
else{

cout << ((len + k)*dx / d)%len << endl;
}
}
return 0;
}

测试数据 #0: Accepted, time = 0 ms, mem = 520 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 528 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 528 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 528 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 528 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 524 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 524 KiB, score = 10

Accepted, time = 0 ms, mem = 528 KiB, score = 100

千万别信那两个用pascal写的，纯粹扯淡
这是AC的
###Block code
#include<cstdio>
using namespace std;
long long m,n,x,y,l,k,s,a,b,d;
void swap(long long &x,long long &y){//{{{
long long t;
t=x,x=y,y=t;
}//}}}
void work(long long a,long long b,long long &x,long long &y){//{{{
if(a==1){
x=1,y=0;
return;
}
long long x1,y1;
work(b%a,a,x1,y1);
y=x1,x=y1-x1*int(b/a);
}//}}}
long long gcd(long long a,long long b){///{{{
if(a==0)return b;
else return gcd(b%a,a);
}//}}}
int main()
{
scanf("%lld%lld%lld%lld%lld",&x,&y,&m,&n,&l);
if(n>m){
swap(m,n);
swap(x,y);
}
m-=n,y-=x,m%=l,y%=l;
if(m==0){
printf("Impossible");
return 0;
}
while(y<0)y+=l;
if(gcd(m,l)!=1){
d=gcd(m,l);
m/=d,l/=d;
if(y%d!=0){
printf("Impossible");
return 0;
}
y/=d;
}
work(m,l,s,k);
while(s<0)s+=l;
s*=y,s%=l;
printf("%lld",s);
return 0;
}

题解

题解

#include<stdio.h>
#include<stdlib.h>
int main()
{
int x,y,m,n,l,i,t=0,tool=0;
scanf("%d %d %d %d %d",&x,&y,&m,&n,&l);
if(m<n)
{
i=0;
while(tool==0)
{
if(x+m*i==y+n*i||i>1000)
{
tool=1;
t=i;
}
else
i++;
}
}
else
{
i=0;
while(tool==0||i>1000)
{
if(x+m>l)
x=x+m-l;
else
x=x+m;
if(y+n>l)
y=x+n-l;
else
y=y+n;
if(x==y)
{
tool=1;
t=i;
}
else
i++;
}
}
if(t!=0)
printf("%d",t);
else
printf("imposible");
return 0;
}

那位大神帮忙看看我这程序那里错了。。。
#include <iostream>
#include <string>
#include <math.h>
#include <time.h>
#include <stdio.h>
using namespace std;
int main()
{
long long x, y, m, n, l;
float days = 0;
float distance = 0;
int i;
cin>>x>>y>>m>>n>>l;
if (x > y) {
distance = x - y;
}
if (x < y) {
distance = y - x;
}
if (x == y)
{
cout<< 0;
return 0;
}
if (m == n)
{
cout<<"Impossible";
return 0;
}

if (m > n)
{
days = distance / (m - n);
}
if (m < n)
{
days = (l - distance) / (n - m);
}
i=days;
if (days > i)
{
cout<<i+1;
}
if (days == i)
{
cout<<i;
}
cout<<endl;
return 0;
}

var

x,y,m,n:1..2000000000;

l:1..2100000000;

begin

read(x,y,m,n,l);

if m=n then write('Imppossible')

else if m>n then write(abs(x-y)div(m-n))

else write((l-abs(x-y))div(n-m));

end.

var

x,y,m,n:1..2000000000;

l:1..2100000000;

begin

read(x,y,m,n,l);

if m=n then write('Imppossible')

else if m>n then write(abs(x-y)div(m-n))

else write((l-abs(x-y))div(n-m));

end.

本人看不懂,于是自己编了一个自己推算的程序

a*k=b( mod c)

---|-->a*k=b+c*l---|-->c*l=-b+a*k---|->c*l=-b ( mod a)

把原方程转化为规模较小的方程，递归求解。。。。

边界条件：b mod a=0 时k= b div a;a=0时无解

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

int64真不是一般的变态.除了读入和赋值的BUG外,还有就是 mod 负数也会出错.比如 int64 类型下,-1 mod -5 = 1.

庆祝100题。。。。。。

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms