POJ-1017 Packets

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题目大意：有货物若干，大小分别为1*1，2*2，3*3，4*4，5*5，6*6，将货物装在6*6大小的箱子中，求最少的箱子个数。
题目链接：http://poj.org/problem?id=1017
简单贪心，从大货物开始考虑即可。

#include <stdio.h>#include <math.h>#include <string.h>int x[7];bool end;int main(){    while (1)    {        end = true;        for (int i = 1; i <= 6; i++)        {            scanf("%d",&x[i]);            if (x[i] != 0)            {                end = false;            }        }        if (end)            break;        int pNum=0;        int tNum = 0;        pNum = x[6];        x[6] = 0;        tNum += pNum;        pNum = x[5];        x[5] = 0;        if (pNum * 11 > x[1])            x[1] = 0;        else            x[1] = x[1] - pNum * 11;        tNum += pNum;        pNum = x[4];        x[4] = 0;        if (pNum * 5 > x[2])        {            if ((pNum * 5 - x[2]) * 4 > x[1])                x[1] = 0;            else                x[1] = x[1] - (pNum * 5 - x[2]) * 4;            x[2] = 0;        }        else        {            x[2] = x[2] - pNum * 5;        }        tNum += pNum;        pNum = x[3] / 4;        if (x[3] % 4 != 0)        {            pNum++;            if ((4 - x[3] % 4) * 2 - 1 > x[2])            {                if (3 * 3 * (4 - x[3] % 4) - 2 * 2 * x[2] > x[1])                    x[1] = 0;                else                    x[1] = x[1] - (3 * 3 * (4 - x[3] % 4) - 2 * 2 * x[2]);                x[2] = 0;            }            else            {                if (3 * 3 * (4 - x[3] % 4) - 2 * 2 * ((4 - x[3] % 4) * 2 - 1) > x[1])                    x[1] = 0;                else                    x[1] = x[1] - (3 * 3 * (4 - x[3] % 4) - 2 * 2 * ((4 - x[3] % 4) * 2 - 1));                x[2] = x[2] - ((4 - x[3] % 4) * 2 - 1);            }        }        x[3] = 0;        tNum += pNum;        pNum = x[2] / 9;        if (x[2] % 9 != 0)        {            pNum++;            if ((9 - x[2] % 9) * 4 > x[1])                x[1] = 0;            else                x[1] = x[1] - (9 - x[2] % 9) * 4;        }        x[2] = 0;        tNum += pNum;        pNum = x[1] / 36;        if (x[1] % 36 != 0)        {            pNum++;        }        x[1] = 0;        tNum += pNum;        printf("%d\n",tNum);    }    return 0;}

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