Created
August 18, 2011 21:31
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| #include <stdio.h> | |
| #include <string.h> | |
| #include <assert.h> | |
| #include <math.h> | |
| #define MAXN 300 | |
| #define min(a,b) ( (a<b)?(a):(b) ) | |
| #define max(a,b) ( (a>b)?(a):(b) ) | |
| int solve(int n) { | |
| int T[MAXN]; | |
| assert(n>=0); | |
| memset(T,0,MAXN*sizeof(int)); | |
| int i,j,k; | |
| int sol = -1; | |
| T[0] = 0; | |
| T[1] = 1; | |
| /*for(k=2;k<=((int)floor(n/2))+1;k++) {*/ | |
| for(k=2;k<=n;k++) { | |
| // compute T[k] | |
| int Tmax = -1; | |
| for(i=1;i<=k/2;i++) { | |
| Tmax = max( | |
| Tmax, | |
| max( | |
| T[i]*T[k-i], | |
| i*(k-i) | |
| ) | |
| ); | |
| }; | |
| T[k] = Tmax; | |
| }; | |
| return T[n]; | |
| } | |
| int main(char argc,char **argv) { | |
| int n; | |
| scanf("%d",&n); | |
| printf("%d\n",solve(n)); | |
| return 0; | |
| }; | |
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