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Problem's Link: http://acm.hnu.cn/online/?action=problem&type=show&id=13412&courseid=0
Mean:
将N分为D份，每份不超过X，有多少种分法？
analyse:
首先我们想到的是迭代，但是数据太大，一路迭代下去必定爆栈+超内存+TLE。
我们枚举X，对于满足条件的X，求和统计答案，不满足条件的X，更新往下迭代的P值。最后对P求和即为答案。
这题DP也可以做，不过上面的方法从时间和空间上都大大优于DP。
Time complexity: O(N)
Source code:
/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-16-16.39
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const LL mod = 1000000007;
LL inv[5000];
LL N,X,D;
void pre()
{
inv[1] = 1;
for(int i=2; i<5000; i++)
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
}
int main()
{
pre();
while(scanf("%d %lld %d",&N,&D,&X) && N)
{
LL ans = 0;
for(int i=0; i*X<=N; i++)
{
LL p = 1;
if(i <= D)
{
for(int j=1; j<=i; j++)
{
p = (D - j + 1) % mod * p % mod;
p = p * inv[j] % mod;
}
}
else p = 0;
for(int j=0; j<i; j++) p = (mod - p);
int gap = N - i*X;
for(int j=1; j<=gap; j++)
{
p = (D + gap - j + mod) % mod * p % mod;
p = p * inv[j] % mod;
}
ans = ans + p;
if(ans >= mod)
ans -= mod;
}
printf("%lld\n",ans);
}
return 0;
}
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-16-16.39
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const LL mod = 1000000007;
LL inv[5000];
LL N,X,D;
void pre()
{
inv[1] = 1;
for(int i=2; i<5000; i++)
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
}
int main()
{
pre();
while(scanf("%d %lld %d",&N,&D,&X) && N)
{
LL ans = 0;
for(int i=0; i*X<=N; i++)
{
LL p = 1;
if(i <= D)
{
for(int j=1; j<=i; j++)
{
p = (D - j + 1) % mod * p % mod;
p = p * inv[j] % mod;
}
}
else p = 0;
for(int j=0; j<i; j++) p = (mod - p);
int gap = N - i*X;
for(int j=1; j<=gap; j++)
{
p = (D + gap - j + mod) % mod * p % mod;
p = p * inv[j] % mod;
}
ans = ans + p;
if(ans >= mod)
ans -= mod;
}
printf("%lld\n",ans);
}
return 0;
}
