### MarioYC's blog

By MarioYC, 6 years ago, ,

Hi everyone, today I solved this problem using Dynamic Programming, the state is easy to get (n,c1,c2,c3) and a useful observation is to notice that if you have n,c1 and c2 then c3 can be found, so it isn't necessary for it to be part of the state. With this and some prunning it is enough to get AC (I though I would get TLE). I see solutions that are much faster, how can this be done?

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define MOD 100000000000000000LL

int r[125];
long long memo[126][126][126];

long long solve(int pos, int c1, int c2, int c3){
if(pos == -1){
if(c1 || c2 || c3) return 0;
return 1;
}

long long &ret = memo[pos][c1][c2];

if(ret == -1){
ret = 0;

for(int i = 0;i <= c1;++i){
for(int j = max(0,r[pos] - i - c3);j <= c2 && i + j <= r[pos];++j){
int k = r[pos] - i - j;

ret += solve(pos - 1,c1 - i,c2 - j,c3 - k);
if(ret >= MOD) ret -= MOD;
}
}
}

return ret;
}

int main(){
int n,c1,c2,c3;

scanf("%d %d %d %d",&n,&c1,&c2,&c3);

int diff = c1 + c2 + c3;

for(int i = 0;i < n;++i){
scanf("%d",&r[i]);
diff -= r[i];
}

if(diff != 0) printf("0\n");
else{
memset(memo,-1,sizeof memo);
printf("%lld\n",solve(n - 1,c1,c2,c3));
}

return 0;
}


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 » 6 years ago, # |   0