zoj 2042
2017-07-07 03:15
211 查看
opt[i,j]表示前i个数%k是否可以得到余数为j的结果。
#include<cstdio>
#include<string.h>
#include<iostream>
#include<cmath>
#define clr(a) memset(a,0,sizeof(a))
#define N 10001
#define M 101
using namespace std;
int a
;
int b
[M];
int p,q;
int n,m;
int opt
[M];
int main()
{
int i,j,k,T,flag;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
clr(b);
clr(opt);
p=1;
b[0][0]=0;
opt[0][0]=1;
for(i=1;i<=n;i++)
{
q=0;
for(j=0;j<p;j++)
{
k=abs(b[i-1][j]+a[i])%m;
if(!opt[i][k])
{
opt[i][k]=1;
b[i][q++]=k;
}
k=abs(b[i-1][j]-a[i])%m;
if(!opt[i][k])
{
opt[i][k]=1;
b[i][q++]=k;
}
}
p=q;
}
if(opt
[0])
puts("Divisible");
else
puts("Not divisible");
if(T)
puts("");
}
return 0;
}
#include<cstdio>
#include<string.h>
#include<iostream>
#include<cmath>
#define clr(a) memset(a,0,sizeof(a))
#define N 10001
#define M 101
using namespace std;
int a
;
int b
[M];
int p,q;
int n,m;
int opt
[M];
int main()
{
int i,j,k,T,flag;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
clr(b);
clr(opt);
p=1;
b[0][0]=0;
opt[0][0]=1;
for(i=1;i<=n;i++)
{
q=0;
for(j=0;j<p;j++)
{
k=abs(b[i-1][j]+a[i])%m;
if(!opt[i][k])
{
opt[i][k]=1;
b[i][q++]=k;
}
k=abs(b[i-1][j]-a[i])%m;
if(!opt[i][k])
{
opt[i][k]=1;
b[i][q++]=k;
}
}
p=q;
}
if(opt
[0])
puts("Divisible");
else
puts("Not divisible");
if(T)
puts("");
}
return 0;
}
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