题解 | TheFewestCoins-牛客假日团队赛2L题

Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the number of coins he uses to pay plus the number of coins he receives in change is minimized. Help him to determine what this minimum number is.
FJ wants to buy cents of supplies. The currency system has different coins, with values . Farmer John is carrying C1 coins of value coins of value coins of value . The shopkeeper has an unlimited supply of all the coins, and always makes change in the most efficient manner (although Farmer John must be sure to pay in a way that makes it possible to make the correct change).

其余几位大佬的题解中说到，最大应一直找到。这是先从店主的角度考虑的

设货币面值为那么当老板需要找的钱时，至少需要枚硬币。设一个前缀和数组为，其***有个元素。根据抽屉原理，至少有两个前缀和与同余，则它们的差。显然当的数量足够时，将这一部分替换成个面值为是最优的（不要纠结真正采用的方案）。对于，同理。
所以在的范围内一定可以找出最优方案，对于更大的也一定可以

顺手贴一下代码，同时感谢题解区的大佬们

#include <cstdio>
#include <cstring>
const int maxT=10000+10;
const int maxn=100+5;
const int maxv=120;
int f[maxT+maxv*maxv],g[maxT+maxv*maxv];//f[i]、g[i]分别表示John和店长付i元钱最少需要用的硬币 
int v[maxn],c[maxn];//如题意所示 
inline int max(int x,int y) {return x>y?x:y;}
inline int min(int x,int y) {return x<y?x:y;}
int main()
{
    int n,t;
    scanf("%d%d",&n,&t);
    for (int i=1;i<=n;i++)
        scanf("%d",&v[i]);
    int sum=0,mx=0;
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&c[i]);
        sum+=c[i]*v[i];
        mx=max(mx, v[i]*v[i]);
    }
    if (sum<t)//买不起，退了 
    {
        printf("-1\n");
        return 0;
    }
    memset(g, 0x3f, sizeof(g));
    memset(f, 0x3f, sizeof(f));
    g[0]=0;
    f[0]=0;
    for (int i=1;i<=n;i++)
        for (int j=v[i];j<=mx;j++)//Rob找j元钱所用的最小钱数 
            g[j]=min(g[j], g[j-v[i]]+1);
    //实际上应该是g[i][j]=min(g[i-1][j], g[i][j-v[i]+1])
    //g[i][j]表示考虑到第i个物品支付j元的最少硬币数 
    //但是因为第一维存储的信息用不到
    //且更新前g[i][j]记录的就是g[i-1][j]的信息 
    //所以可以只用一维 
    for (int i=1;i<=n;i++)
    { 
        for (int j=1;j<=c[i];j<<=1)
        {
            for (int k=t+mx;k>=j*v[i];k--)//倒过来更新（实际上是拆分成01背包的形式） 
                f[k]=min(f[k], f[k-j*v[i]]+j);
            c[i]-=j;//相当于用拆分的物品进行了一次更新，要从数量中减去 
        }
        if (c[i])//还有剩余的没更新 
            for (int k=t+mx;k>=c[i]*v[i];k--)
                f[k]=min(f[k], f[k-c[i]*v[i]]+c[i]);
    }
    int ans=0x3f3f3f3f;
    for (int i=t;i<=t+mx;i++)
        ans=min(ans, f[i]+g[i-t]);
    printf("%d\n",ans==0x3f3f3f3f?-1:ans);
    return 0;
}