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Use when the code branches on equality of input parameters and returns early via specialized closed-form computations (e.g., Pow) for degenerate cases
Codex 또는 Claude로 설치 이 Prompt를 복사해 Codex, Claude 또는 다른 어시스턴트에 붙여 넣으면 Skill 페이지를 검토하고 설치를 진행할 수 있습니다.
SOC 직업 분류 기준
Use when printf statements output strings without trailing newlines
Use when array is accessed with direct user input as index (0-based indexing)
Use when the program accesses a precomputed table using an offset (e.g., arr[n-1]) that introduces an arithmetic operation between the symbolic input and the index
Use when loop variables iterate over character values (e.g., ASCII codes) to represent different operations or choices
Use when code uses a loop to search for an input value in an array and then uses the found index for further computation
Use when code reads a fixed number of characters into an array and only accesses individual elements
| name | avoid-special-case-early-returns |
| description | Use when the code branches on equality of input parameters and returns early via specialized closed-form computations (e.g., Pow) for degenerate cases |
A KLEE-coverage code transformation. Applying it rewrites C source so symbolic execution explores more of the program's behavior.
When the code branches on equality of input parameters and returns early via specialized closed-form computations (e.g., Pow) for degenerate cases
Remove special-case shortcut branches and compute the result through the unified general DP/iterative formulation that naturally handles all input combinations
Early-return branches based on input equality create disjoint path partitions that KLEE must explore separately, and the closed-form Pow loop introduces input-dependent loop bounds that complicate symbolic constraints; a single unified computation yields fewer paths and more uniform constraints, letting generated tests cover more behaviors when replayed
Before:
#include<stdio.h>
#pragma GCC optimize ("O2")
#define rep(i,N) for(int i=0;i<(int)N;i++)
const int MOD=998244353;
int Pow(int x,int y){int z=1;x%=MOD;while(y){if(y&1){z=(1ll*z*x)%MOD;}x=(1ll*x*x)%MOD;y>>=1;}return z;}
int main(void)
{
int A,B,C,D,ans=0,sum,dp[3000]={0};
if(!scanf("%d %d %d %d",&A,&B,&C,&D))return 0;
if(A==C){return printf("%d\n",Pow(A,D-B)),0;}
else if(B==D){return printf("%d\n",Pow(B,C-A)),0;}
dp[A-1]=1;
rep(i,D-B)
{
sum=0;
rep(j,C)
{
sum=(1ll*sum*(B+i)%MOD+dp[j])%MOD;
dp[j]=(1ll*j*dp[j]+sum)%MOD;
}
}
rep(i,C){ans=(ans+1ll*dp[i]*Pow(D,C-i-1))%MOD;}
return printf("%d\n",ans),0;
}
After:
#include <stdio.h>
#define MOD 998244353
long long int DP[3010][3010];
int main()
{
int a,b,c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
long long int C = 1;
for(int j=c;j>=a;j--)
{
DP[d][j] = C;
C*=d;
C%=MOD;
}
for(int i=d-1;i>=b;i--)
{
C = 0;
for(int j=c;j>=a;j--)
{
DP[i][j] = (C + j*DP[i+1][j])%MOD;
//C *= i;
C += DP[i+1][j];
C *= i;
C %= MOD;
//printf("%d %d %lld!!\n",i,j,DP[i][j]);
}
}
printf("%lld",DP[b][a]);
}
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