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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
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Baseado na classificação ocupacional SOC
| 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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