算法:用前缀和把O(n^3)的dp优化成O(n^2)。注意:(1)long long。(2)L1、L2不一定在[1,300]内。(3)没有空行(不用处理空行能过)。
//10313 Pay the Price Accepted C++ 0.056 2012-11-20 11:23:14
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 300+5;
const int MAX = 300;
int N, L1, L2;
long long d[MAXN][MAXN], f[MAXN][MAXN];
char line[100];
int main()
{
d[0][0] = f[0][0] = 1;
for (int i = 1; i <= MAX; i++)
for (int j = 1; j <= i; j++)
{
d[i][j] = f[i-j][min(j, i-j)];
f[i][j] = f[i][j-1]+d[i][j];
}
while (gets(line))
{
int n = sscanf(line, "%d%d%d", &N, &L1, &L2);
if (n > 0)
{
if (n == 1)
printf("%lld\n", f[N][N]);
else if (n == 2)
printf("%lld\n", f[N][min(L1, N)]);
else if (L1 > N)
printf("0\n");
else if (L1 > 0)
printf("%lld\n", f[N][min(L2, N)]-f[N][L1-1]);
else
printf("%lld\n", f[N][min(L2, N)]);
}
}
return 0;
}