hdu 4472 Count (2012 ACM-ICPC 成都现场赛)
2013-09-01 17:12
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递推,考虑到一n可以由i * j + 1组合出来,即第二层有j个含有i个元素的子树。。。然后就可以了。。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<fstream>
#include<sstream>
#include<bitset>
#include<vector>
#include<string>
#include<cstdio>
#include<cmath>
#include<stack>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
#define SL(a) strlen(a)
using namespace std;
const int N = 1111;
const int MOD = 1e9 + 7;
LL ans
;
int main()
{
int n, cas = 1, i, j;
CLR(ans, 0);
ans[1] = 1;
for(i = 1; i < N; i ++)
{
for(j = 1; i * j + 1 < N; j ++)
{
ans[j * i + 1] += ans[i];
ans[j * i + 1] %= MOD;
}
}
while(cin >> n)
{
cout << "Case " << cas ++ << ": ";
cout << ans
<< endl;
}
}
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<fstream>
#include<sstream>
#include<bitset>
#include<vector>
#include<string>
#include<cstdio>
#include<cmath>
#include<stack>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
#define SL(a) strlen(a)
using namespace std;
const int N = 1111;
const int MOD = 1e9 + 7;
LL ans
;
int main()
{
int n, cas = 1, i, j;
CLR(ans, 0);
ans[1] = 1;
for(i = 1; i < N; i ++)
{
for(j = 1; i * j + 1 < N; j ++)
{
ans[j * i + 1] += ans[i];
ans[j * i + 1] %= MOD;
}
}
while(cin >> n)
{
cout << "Case " << cas ++ << ": ";
cout << ans
<< endl;
}
}
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