原理简单..N+1个物品放入N个盒子中..必定有一个盒子里物品大于一个...
但要知道什么时候用...1、may print any of them ..Special judge...就像这两题..如果要输出数最少的..估计DP了.
2、有类似取模的运算...
Program ( POJ 2356 )
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#define ll long long
#define eps 1e-5
#define oo 10007
#define pi acos(-1.0)
#define MAXN 10005
using namespace std;
int sum[MAXN],a[MAXN],times[MAXN];
int main()
{
int i,j,x,n;
sum[0]=0;
memset(times,0,sizeof(times));
scanf("%d",&n);
for (i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum[i]=(sum[i-1]+a[i])%n;
times[sum[i]]++;
}
if (times[0])
{
for (i=1;i<=n;i++)
if (sum[i]==0) break;
printf("%d\n",i);
for (x=1;x<=i;x++) printf("%d\n",a[x]);
}else
{
for (x=1;x<n;x++)
if (times[x]>=2) break;
for (i=1;i<=n;i++)
if (sum[i]==x) break;
for (j=i+1;j<=n;j++)
if (sum[j]==x) break;
printf("%d\n",j-i);
for (x=i+1;x<=j;x++) printf("%d\n",a[x]);
}
return 0;
}
Program ( POJ 3370 )
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#define ll long long
#define eps 1e-5
#define oo 10007
#define pi acos(-1.0)
#define MAXN 100005
using namespace std;
int sum[MAXN],a[MAXN],times[MAXN];
int main()
{
int i,j,x,c,n;
while (~scanf("%d%d",&c,&n) && n && c)
{
sum[0]=0;
memset(times,0,sizeof(times));
for (i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum[i]=(sum[i-1]+a[i])%c;
times[sum[i]]++;
}
if (times[0])
{
for (i=1;i<=n;i++)
if (sum[i]==0) break;
for (x=1;x<i;x++) printf("%d ",x);
printf("%d",i);
}else
{
for (x=1;x<c;x++)
if (times[x]>=2) break;
for (i=1;i<=n;i++)
if (sum[i]==x) break;
for (j=i+1;j<=n;j++)
if (sum[j]==x) break;
for (x=i+1;x<j;x++) printf("%d ",x);
printf("%d",j);
}
printf("\n");
}
return 0;
}
