POJ 1017 Packets

Test instructions: There are some 1x1, 2x2, 3x3, 4x4, 5x5, 6x6 cargo, each cargo height is H, the goods are packed, each package can be loaded 6x6xh, ask at least a few parcels.

Solution: 6x6 directly into the 5x5 space can be filled with 1x1, 4x4 can be used 2x2 and 1x1 fill, 3x3 Four can be composed of a package, a multi-and 2x2 and 1x1 combination, 2x2 9 composed of a parcel, multi and 1x1 combination, 1x 1 of the 36 to form a package, in order to simulate the above Otz do not know how to think ... Write a lot of bugs to

Code:

#include <stdio.h> #include <iostream> #include <algorithm> #include <string> #include < string.h> #include <math.h> #include <limits.h> #include <time.h> #include <stdlib.h># include<map> #include <queue> #include <set> #include <stack> #include <vector> #define LL    Long longusing namespace Std;int main () {int A, B, C, D, E, F; while (~SCANF ("%d%d%d%d%d%d", &a, &b, &c, &d, &e, &f) &&! ( A = = 0 && b = 0 && c = 0 && D = = 0 && e = 0 && f = 0) {int ans = 0        ;        Ans + = f;        Ans + = e;        int TMP = one * e;        A = max (0, a-tmp);        ans + = D;        TMP = 5 * D; if (b < tmp) {tmp-= b;//because these two sentences have been reversed for a half-day bug ...        The IQ of the clumsy B = 0;            } else {b = tmp;        TMP = 0;        } a = max (0, a-tmp * 4);        Ans + = C/4;        C%= 4;        TMP = 0; if (c)       {ans++;            TMP = 36-c * 9;            int ttmp = 0;            if (c = = 1) ttmp = 5;            else if (c = = 2) ttmp = 3;            else ttmp = 1;            TMP-= ttmp * 4;                if (b < ttmp) {ttmp-= b;            b = 0;                } else {b-= ttmp;            ttmp = 0;        } a = max (0, a-ttmp * 4-tmp);        } ans + = B/9;        b%= 9;            if (b) {ans++;        A = max (0, a-36 + 4 * b);        } ans + = A/36;        A%= 36;        if (a) ans++;    printf ("%d\n", ans); } return 0;}

　　

POJ 1017 Packets