51nod1057 N的阶乘

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链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1057

题意：中文题。

分析：裸的高精度会T，压压位就过了。

代码（压4位）：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=50005;const int mod=100000000;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=998244353;const int INF=1000000010;const ll MAX=1000000000000;const double pi=acos(-1.0);typedef double db;typedef unsigned long long ull;int a[N];int main(){    int i,j,n;    scanf("%d", &n);    a[0]=a[1]=1;    for (i=2;i<=n;i++) {        for (j=1;j<=a[0];j++) a[j]*=i;        for (j=1;j<a[0];j++) a[j+1]+=a[j]/10000,a[j]%=10000;        while (a[a[0]]>9999) {            a[a[0]+1]+=a[a[0]]/10000;            a[a[0]]%=10000;a[0]++;        }    }    printf("%d", a[a[0]]);    for (i=a[0]-1;i;i--)    if (a[i]<10) printf("000%d", a[i]);    else if (a[i]<100) printf("00%d", a[i]);        else if (a[i]<1000) printf("0%d", a[i]);            else printf("%d", a[i]);    printf("\n");    return 0;<h3>}</h3>

代码（压8位）：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=50005;const int mod=100000000;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=998244353;const int INF=1000000010;const ll MAX=1000000000000;const double pi=acos(-1.0);typedef double db;typedef unsigned long long ull;ll a[N];int main(){    int j,n;    scanf("%d", &n);    a[0]=a[1]=1ll;    for (ll i=2;i<=n;i++) {        for (j=1;j<=a[0];j++) a[j]*=i;        for (j=1;j<a[0];j++) a[j+1]+=a[j]/100000000ll,a[j]%=100000000ll;        while (a[a[0]]>99999999ll) {            a[a[0]+1]+=a[a[0]]/100000000ll;            a[a[0]]%=100000000ll;a[0]++;        }    }    printf("%d", a[a[0]]);    for (j=a[0]-1;j;j--)    if (a[j]<10ll) printf("0000000%I64d", a[j]);    else if (a[j]<100ll) printf("000000%I64d", a[j]);        else if (a[j]<1000ll) printf("00000%I64d", a[j]);            else if (a[j]<10000ll) printf("0000%I64d", a[j]);                else if (a[j]<100000ll) printf("000%I64d", a[j]);                    else if (a[j]<1000000ll) printf("00%I64d", a[j]);                        else if (a[j]<10000000ll) printf("0%I64d", a[j]);                            else printf("%I64d", a[j]);    printf("\n");    return 0;}

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