# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
22596 | 크리콘 B번 문제는 그리디로 풀려요 (#40) | Unifying Values (KRIII5_UV) | C++98 | 500 ms | 3732 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include<stdio.h>
#include<map>
#define ab(a) ((a)<0?-(a):(a))
using namespace std;
typedef long long lld;
const lld mod = 1000000007;
int N, scn;
lld ba[10101], sum[10101], dap;
map<pair<lld,int>, lld> dp;
lld dy(lld sm, int p){
pair<lld,int> pp = make_pair(sm, p);
if(p>N)return 1;
if(dp.find(pp) != dp.end()) return dp[pp];
lld dap=0;
for(int i=p; i<=N; i++){
if(sum[i]-sum[p-1] == sm) dap += dy(sm,i+1);
}
return dp[pp] = dap%mod;
}
int main(){
scanf("%d", &N);
for(int i=1; i<=N; i++){
scanf("%lld", &ba[i]), sum[i]=sum[i-1]+ba[i];
}
for(int i=1; i<=N; i++){
if(sum[N] == 0){
if(sum[i] == 0) dap += dy(0, i+1);
}
else{
if(sum[i] == 0)continue;
if(ab(sum[N]) % ab(sum[i]))continue;
if(sum[N] / sum[i] <= 0)continue;
dap += dy(sum[i], i+1);
}
}
printf("%lld", (dap+mod-1)%mod);
return 0;
}
Compilation message (stderr)
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