HDU 1005 Number Sequence
2014-08-14 10:06
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题解:因为模比较小,所以一定会产生循环节,所有先计算循环节,然后直接求解。
#include <cstdio> int main(){ int a,b,n,f[50]; f[1]=f[2]=1; while(scanf("%d%d%d",&a,&b,&n),a|b|n){ int t1,t2,bo=0; for(int i=3;i<=n&&!bo;i++){ f[i]=(a*f[i-1]+b*f[i-2])%7; for(int j=2;j<=i-1;j++)if(f[i]==f[j]&&f[i-1]==f[j-1]){t1=j,t2=i;bo=1;break;} }if(bo)printf("%d\n",f[t1+(n-t2)%(t2-t1)]);else printf("%d\n",f ); } return 0; }
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