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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1509 Accepted Submission(s): 498
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <map>
#include <bitset>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
#define CT continue
#define SC scanf
const int N=2*1e5+10;
int factor[N],num[N],appear[N],step[N];
ll add[N];
int gcd(int a,int b)
{
if(b==0) return a;
else return gcd(b,a%b);
}
int main()
{
int cas,n,m,kk=0;
SC("%d",&cas);
while(cas--){
SC("%d%d",&n,&m);
int cnt=0;MM(num,0);MM(appear,0);
for(int i=1;i<=n;i++) SC("%d",&step[i]);
for(int i=1;i*i<=m;i++) if(m%i==0){
factor[++cnt]=i;
if(i*i!=m) factor[++cnt]=m/i;
}
sort(factor+1,factor+cnt+1);
cnt--;
for(int i=1;i<=cnt;i++){
ll k=(m-1)/factor[i];
add[i]=k*(k+1)/2*factor[i];
}
for(int i=1;i<=n;i++){
int k=gcd(step[i],m);
for(int j=1;j<=cnt;j++)
if(factor[j]%k==0) appear[j]=1;
}
ll ans=0;
for(int i=1;i<=cnt;i++) if(num[i]!=appear[i]){
ans+=add[i]*(appear[i]-num[i]);
for(int j=i+1;j<=cnt;j++)
if(factor[j]%factor[i]==0) num[j]+=(appear[i]-num[i]);
}
printf("Case #%d: %lld\n",++kk,ans);
}
return 0;
}
分析:
1.一个数的因子个数大概是log级别;
2.ax+by=c有非负整数解的条件是c%gcd(a,b);取余
=>ax+by=k*gcd(a,b) =>ax%b=k*gcd(a,b)%b =>ai*x%m=k*gcd(ai,m)%m;
所以,青蛙能走到的格子数是k*gcd(ai,m),而gcd(ai,m)又必然是m的因数,所以可以先分解出m的因子
3.但是因为有些格子可能同时被多只青蛙走,因此需要容斥一下,设appear[i]为格子i应该走的次数(只有0,1)两个值,num[i]为格子i到当前为止实际走的次数,如果当前appear[i]>appear[i],就需要加,否则减去,然后将是当前因子factor[i]倍数的因子也同时跟新num值
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原文地址:http://www.cnblogs.com/smilesundream/p/5899059.html