求N^N最左边的一位
1.转化为小数,快速幂
2.数学公式
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#include<cstdio> #include<cmath> using namespace std; long long n; double n0; int cal(long long x) { return (int)log10(x); } int ksm(double a,long long b) { double tmp=a,ret=1.0; while(b) { if(b & 1) { ret*=tmp; if(ret>10)ret/=10; } tmp*=tmp; if(tmp>10)tmp/=10; b>>=1; } return (int)ret; } int main() { int t; scanf("%d",&t); while(t--) { scanf("%lld",&n); n0=(double)n/pow(10.0,cal(n)); int ans=ksm(n0,n); printf("%d ",ans); } return 0; }
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#include<cstdio> #include<cmath> using namespace std; int cal(long long x) { long long tmp1=(long long)(log10(x)*x); double tmp2=x*log10(x); double tmp=tmp2-tmp1; int ret=(int)pow(10.0,tmp); return ret; } int main() { int t; scanf("%d",&t); while(t--) { long long n; scanf("%lld",&n); int ans=cal(n); printf("%d ",ans); } return 0; }
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