C 解法, 执行用时: 2ms, 内存消耗: 304KB, 提交时间: 2021-08-29

#include<stdio.h>
int main()
{
    long int a[2][7];
    long int i,num=0;
    for(i=0;i<7;i++)
    {
        scanf("%ld ",&a[0][i]);
    }
    a[1][0]=1,a[1][1]=5,a[1][2]=10,a[1][3]=50,a[1][4]=100,a[1][5]=500,a[1][6]=a[0][6];
    for(i=5;i>=0;i--)
    {
        if(a[0][i]*a[1][i]>=a[0][6])
        {
            num=num+a[0][6]/a[1][i];
            a[0][6]=a[0][6]%a[1][i];
        }
        else
        {
            a[0][6]=a[0][6]-a[0][i]*a[1][i];
            num=num+a[0][i];
        }
        if(a[0][6]==0) break;
    }
    if(a[0][6]==0) printf("%ld\n",num);
    else printf("NOWAY");
}

C 解法, 执行用时: 2ms, 内存消耗: 356KB, 提交时间: 2019-04-01

#include<stdio.h>
int main()
{
    int c1,c5,c10,c50,c100,c500,a,b,c,d,e,f,g,zz;
    scanf("%d %d %d %d %d %d %d",&c1,&c5,&c10,&c50,&c100,&c500,&a);
        b=a/500;
        a=a-500*b;
        c=a/100;
        a=a-100*c;
        d=a/50;
        a=a-50*d;
        e=a/10;
        a=a-10*e;
        f=a/5;
        a=a-5*f;
        g=a/1;
        if((g<=c1)&&(f<=c5)&&(e<=c10)&&(d<=c50)&&(c<=c100)&&(b<=c500))
            {zz=g+f+e+d+c+b;printf("%d",zz);}
        else printf("NOWAY");
}

C 解法, 执行用时: 2ms, 内存消耗: 368KB, 提交时间: 2021-10-31

#include <stdio.h>
#include <stdlib.h>
 
int main()
{
    int A,C1,C2,C3,C4,C5,C6,a,b,c,d,e,f;
    int sum;
    scanf("%d %d %d %d %d %d %d",&C1,&C2,&C3,&C4,&C5,&C6,&A);
    e = A/500;
    A = A%500;
    d = A/100;
    A = A%100;
    c = A/50;
    A = A%50;
    b = A/10;
    A = A%10;
    a = A/5;
    A = A%5;
    f = A;
    if((f <= C1)&&(a <= C2)&&(b <= C3)&&(c <= C4)&&(d <= C5)&&(e <= C6))
    {
        sum = a+b+c+d+e+f;
        printf("%d\n",sum);
    }
    else
        printf("NOWAY");
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 368KB, 提交时间: 2020-09-27

#include <stdio.h>

int i=6;
int a[6]={0},b[6]={1, 5, 10, 50, 100, 500};
int sum=0;
int fun(int x)
{
    i -= 1;
    if(x%b[i]==0 && x/b[i]<=a[i]) return sum+x/b[i];
    else if(x%b[i]==0 && x/b[i]>a[i]) return -1;
    else if(x/b[i]<=a[i])
    {
        sum += x/b[i];
        x = x-x/b[i]*b[i];
        return fun(x);
    }
    else
    {
        sum += a[i];
        x = x-x/b[i]*a[i];
        return fun(x);
    }
}
int main()
{
    int A;
    scanf("%d%d%d%d%d%d",&a[0],&a[1],&a[2],&a[3],&a[4],&a[5]);
    scanf("%d", &A);
    int n;
    n = fun(A);
    if(n == -1) printf("NOWAY\n");
    else printf("%d\n", n);
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 368KB, 提交时间: 2019-04-10

#include <stdio.h>
#include <stdlib.h>
 
int main()
{
    int A,C1,C2,C3,C4,C5,C6,a,b,c,d,e,f;
    int sum;
    scanf("%d %d %d %d %d %d %d",&C1,&C2,&C3,&C4,&C5,&C6,&A);
    e = A/500;
    A = A%500;
    d = A/100;
    A = A%100;
    c = A/50;
    A = A%50;
    b = A/10;
    A = A%10;
    a = A/5;
    A = A%5;
    f = A;
    if((f <= C1)&&(a <= C2)&&(b <= C3)&&(c <= C4)&&(d <= C5)&&(e <= C6))
    {
        sum = a+b+c+d+e+f;
        printf("%d\n",sum);
    }
    else
        printf("NOWAY");
    return 0;

}