# 2015ACM/ICPC亚洲区沈阳站-重现赛 HUD 5514 Frogs (容斥原理+GCD) ## Frogs

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2828 Accepted Submission(s): 909

Problem Description

There are

m

stones lying on a circle, and

n

frogs are jumping over them.
The stones are numbered from

0

to

m

1

and the frogs are numbered from

1

to

n

. The

i

-th frog can jump over exactly

a

i

stones in a single step, which means from stone

j

m
o

m

to stone

(
j
+

a

i

)

m
o

m

(since all stones lie on a circle).
All frogs start their jump at stone

0

, then each of them can jump as many steps as he wants. A frog will occupy a stone when he reach it, and he will keep jumping to occupy as much stones as possible. A stone is still considered “occupied” after a frog jumped away.
They would like to know which stones can be occupied by at least one of them. Since there may be too many stones, the frogs only want to know the sum of those stones’ identifiers.
Input

There are multiple test cases (no more than

20

), and the first line contains an integer

t

,
meaning the total number of test cases.
For each test case, the first line contains two positive integer

n

and

m

– the number of frogs and stones respectively

(
1

n

10

4

,

1

m

10

9

)

.
The second line contains

n

integers

a

1

,

a

2

,

,

a

n

, where

a

i

denotes step length of the

i

-th frog

(
1

a

i

10

9

)

.
Output
For each test case, you should print first the identifier of the test case and then the sum of all occupied stones’ identifiers.
Sample Input

3 2 12 9 10 3 60 22 33 66 9 96 81 40 48 32 64 16 96 42 72
Sample Output

Case #1: 42 Case #2: 1170 Case #3: 1872

【题意】

【思路】

m的因子为 1 2 3 4 6 8 12

2 3 4 6

2 4 6 8 10 （5个）

3 6 9 ( 3个）

4 8 （2 个）

6 （1 个）

num【】一次为 0 0 1 2 然后 ans += 前x 项和 * （vis【i】— num【i】） （非常巧妙）

【代码实现】

```#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%dn",n)
#define lson rt << 1, l, mid
#define rson rt <>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}
ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}
ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}
ll inv2(ll b){return qpow(b,MOD-2);}

ll n,m;
ll a[maxn];
ll vis[maxn];
ll qoop[maxn];
ll num[maxn];
ll cot;
void init()
{
cot=0;
for(int i=1;i>t;
int cont=0;
while(t--)
{
scanf("%lld %lld",&n,&m);
mem(a,0);
mem(num,0);
mem(vis,0);
mem(qoop,0);
init();
sort(qoop+1,qoop+cot+1);
for(int i=1;i<=n;i++) 2="" {="" ll="" x;="" scanf("%lld",&x);="" temp="gcd(x,m);" for(int="" j="1;j<=cot;j++)" if(qoop[j]%temp="=0)" vis[j]="1;" }="" ans="0;" i="1;i<=cot;i++)" if(vis[i]!="num[i])" tm="(m-1)/" qoop[i];="" ans+="tm" *="" (tm+1)="" *qoop[i]="" (vis[i]-num[i]);="" if(qoop[j]%qoop[i]="=0)" num[j]+="vis[i]-num[i];" printf("case="" #%d:="" %lldn",++cont,ans);="" return="" 0;="" }
123```  5. ### 5G还未商用，但6G的研究早已开始 