20.R的n次冪scanf memset

輸入
The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.

輸出
The output will consist of one line for each line of input giving the exact value of Rn. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don’t print the decimal point if the result is an integer.

樣例
輸入樣例 1 複製
#R n
95.123 12
0.4321 20
5.1234 15
6.7592 9
98.999 10
1.0100 12

#include<stdio.h>
#include<string.h>
#include<math.h>
#define p 7
#define len 120
int main(){
    int mult(int*a,int*b);
    int n,i,j,k,location,flag;
    int an[len],bn[len];
    char sz[p];
    /**scanf掃描n變量返回整數n，注意需要使用&符號，就是直接引用變量，改變變量值**/
    while(scanf("%s %d",sz,&n)==2){
            memset(an,0,sizeof(an));//分配內存，指向an，內存大小爲sizeof(an)，元素初始化爲0的Sas碼
            memset(bn,0,sizeof(bn));
            an[0]=1;
            location=-1;
        for(i=0;i<p-1;i++){
            if(sz[i]=='.'){
                location=i;//memory the location of the dot
                location=n*(p-2-i);//get the new location
                for(j=i;j<p-1;j++)
                    sz[j]=sz[j+1];//remove the dot
                    sz[p-1]='\0';
            }
        }
        for(i=0;i<strlen(sz);i++)
            bn[i]=sz[strlen(sz)-1-i]-'0';
             for(i=0;i<n;i++)
            mult(an,bn);
        if(location==-1){
        flag=0;
        for(i=len-1;i>=0;i--){
            if(flag)printf("%d",an[i]);
        else if(an[i]){
            printf("%d",an[i]);
            flag=1;
            }
        }
         if(!flag)printf("0");
         printf("\n");
        }//整數情況
        else//小數
        {
            for(i=0;;i++){
                 k=i;
                if(an[i]!=0)break;
            }
           for(i=len-1;i>=k;i--){
               if(an[i])break;
           }
           if(i<location){
            printf(".");
            for(i=location-1;i>=k;i--)printf("%d",an[i]);
            printf("\n");
           }
           else{
                if(k<location){
            flag=0;
            for(i=len-1;i>=k;i--){
            if(flag){
                    printf("%d",an[i]);
                    if(i==location&&i!=k)printf(".");
            }
            else if(an[i]){
            printf("%d",an[i]);
            if(i==location&&i!=k)printf(".");
            flag=1;
            }
        }
                    printf("\n");
           }
           else{
            flag=0;
            for(i=len-1;i>=location;i--){
            if(flag){
                    printf("%d",an[i]);
            }
            else if(an[i]){
            printf("%d",an[i]);
            flag=1;
            }
        }
          if(!flag)printf("0");
          printf("\n");
           }
           }
        }
    }
}
int mult(int*a,int*b){
    int i,j,k,l;
    int c[len];
    memset(c,0,sizeof(c));
    for(i=0;i<len;i++){
        k=0;
        for(j=0;j<len-i;j++){
            l=c[i+j]+a[j]*b[i]+k;
            c[i+j]=l%10;
            k=l/10;
        }
    }
    for(i=0;i<len;i++)a[i]=c[i];
}