sdut2146 - 最小子序列和


http://www.voidcn.com/article/p-uujpcruu-pc.html
给你一个长为n(10<=n<=10000)的数组,数组中的每一个数大于等于1小于等于1000000。请你找出一个长为k(100<=k&lt;=1000)的子序列。找序列时,假如第一个数找的是数组中的第i个位置的数,那么找第二个数时只能找数组中第i个位置后的数,依次找出k个数。使得第一个数*1+第二个数*2+...+第k个数*k的值最小。 

输入

 有多组(小于11组)测试数据,每组第一行输入nk(用空格隔开),第二行输入n个数(数之间用空格隔开)。

输出

 请输出最小的和。

示例输入

15 5
5 4 3 2 1 1 2 3 4 5 5 4 3 2 1

示例输出

19
int main()
{
    //freopen("//media/学习/ACM/input.txt","r",stdin);
    while(~scanf("%d%d",&n,&m))
    {
        int i,j;
        for( dp[0][0]=0,i=1;i<=n;i++)scanf("%d",&a[i]),dp[1][i]=dp[0][i]=0;
        for(i=1;i<=m;i++)
        {
            ans=INF;//代表j之前
            for(j=i;j<=n;j++)
            {
                dp[1][j]=dp[0][j-1]+a[j]*i;
                dp[0][j-1]=ans;
                if(ans>dp[1][j])ans=dp[1][j];
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}

https://blog.csdn.net/wyk1823376647/article/details/52511331
int dp[2][100011];
int main()
{
    int i,n,k,pl,j,ans;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        for (i = 1 ;i <= n; i++)
            scanf("%d",&pa[i]);
        memset(dp,0,sizeof(dp));
        for(i = 1; i <= k; i++)
        {
            ans = 0x3f3f3f3f;
            for( j = i ; j <= n ; j++)
            {
                dp[1][j] = dp[0][j-1] + pa[j] * i; //当前最优解 = 在 j 之前的最优解 + pa[j] * i;
                dp[0][j-1] = ans; // 当前 j-1最优解
                if(ans > dp[1][j])
                    ans = dp[1][j];
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}


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