HDU 1268 (a pile of bricks castle)

The minimum number of block columns by calculation, each taking one (Z-axis); Maximum Number of blocks calculated in layers, each covered as much as possible.

(HDU compiler is easy to use and __int64 long long type in C ++ compiler error and difficult investigation, recommended the use of both types, according to the syntax of the C language, the use scanf and printf)

#include <stdio.h>
#include <string.h>

long long front[100005], side[100005]; //正面和侧面每层的木块数
int frontHigh[100005], sideHigh[100005]; //正面和侧面每个高度出现的次数

int main()
{
    int W, L;
    while (~scanf("%d %d", &W, &L))
    {
        memset(front, 0, sizeof(front));
        memset(side, 0, sizeof(side));
        memset(frontHigh, 0, sizeof(frontHigh));
        memset(sideHigh, 0, sizeof(sideHigh));

        int frontMaxH = 0, sideMaxH = 0; //正面和侧面最大高度
        int temp;
        for (int i = 0; i < W; i++) //输入正面
        {
            scanf("%d", &temp);
            frontHigh[temp]++;
            for (int j = 1; j <= temp; j++)
            {
                front[j]++;
            }
            frontMaxH = temp > frontMaxH ? temp : frontMaxH;
        }
        for (int i = 0; i < L; i++) //输入侧面
        {
            scanf("%d", &temp);
            sideHigh[temp]++;
            for (int j = 1; j <= temp; j++)
            {
                side[j]++;
            }
            sideMaxH = temp > sideMaxH ? temp : sideMaxH;
        }

        if (sideMaxH != frontMaxH)
        {
            printf("No solution.\n");
            continue;
        }

        long long M = 0, N = 0; //最少木块数，最多木块数
        for (int i = 1; i <= frontMaxH; i++) //计算最少木块数
        {
            if (frontHigh[i] > 0 || sideHigh[i] > 0)
            { //按各高度的最大数量取木块，按列取
                if (frontHigh[i] > sideHigh[i])
                    M += frontHigh[i] * i;
                else
                    M += sideHigh[i] * i;
            }
        }
        for (int i = 1; i <= frontMaxH; i++) //计算最多木块数
        { //每层填满，为矩形，按层取
            N += front[i] * side[i];
        }
        printf("%lld %lld\n", M, N);
    }
    return 0;
}

Keep up.