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/*Cow Bowling Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8030 Accepted: 5248 Description The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this: 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 Then the other cows traverse the triangle starting from its tip and moving "down" to one of the two diagonally adjacent cows until the "bottom" row is reached. The cow's score is the sum of the numbers of the cows visited along the way. The cow with the highest score wins that frame. Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable. Input Line 1: A single integer, N Lines 2..N+1: Line i+1 contains i space-separated integers that represent row i of the triangle. Output Line 1: The largest sum achievable using the traversal rules Sample Input 5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 Sample Output 30 Hint Explanation of the sample: 7 * 3 8 * 8 1 0 * 2 7 4 4 * 4 5 2 6 5 The highest score is achievable by traversing the cows as shown above. */ #include <stdio.h> #define MAX_COW_NUM 61426/*1+2+...+350 +1*/ #define MAX(a,b) (((a) > (b)) ? (a) : (b)) int gaiScore[MAX_COW_NUM]; int CowBowlingmain(void) { int iLoop; int iLoopCow; int iCowNum = 0; int iRowNum; int iRowCowNum; int iIndex; scanf("%d",&iRowNum); iCowNum = (iRowNum+1)*iRowNum/2; for (iLoopCow = 1; iLoopCow <= iCowNum; iLoopCow++) { scanf("%d",&gaiScore[iLoopCow]); } iRowCowNum = iRowNum-1; iLoop = 0; for (iLoopCow = iCowNum - iRowNum; iLoopCow >= 1; iLoopCow--) { iIndex = iLoopCow+iRowCowNum; gaiScore[iLoopCow] += MAX(gaiScore[iIndex],gaiScore[iIndex+1]); iLoop++; if (iLoop == iRowCowNum) { iLoop = 0; iRowCowNum--; } } printf("%d/n",gaiScore[1]); return 0; }
