ImportJava.util.Scanner; Public classMain { Public Static voidMain (string[] args) {int[] Trinumbers =New int[101] [101];//0-100 inside int[] Trimaxs =New int[101] [101]; Scanner in=NewScanner (system.in); intNrow = In.nextint ();//Number of rows for(inti=0;i<nrow;i++)//Line { for(intj=0;j<=i;j++)//in-line{Trinumbers[i][j]=In.nextint (); }} in.close (); for(inti = nrow-1;i>=0;i--) { for(intj=i;j>=0;j--)//in-line { if(nrow==i) trimaxs[i][j]= Trinumbers[i][j];//last line ElseTrimaxs[i][j]= Math.max (Trimaxs[i+1][j], trimaxs[i+1][j+1]) +Trinumbers[i][j]; }} System.out.println (trimaxs[0][0]); }}
Topic:
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Total time limit:
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1000ms
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Memory Limit:
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65536kB
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Describe
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7
3 8
8 1 0
2 7 4 4
4 5 2) 6 5
(Fig. 1)
Figure 1 shows a number triangle. There are many different paths from the top of the triangle to the bottom. For each path, add up the number of paths above to get one and your task is to find the largest and.
Note: Each step on the path can only go from one number to the next, or to the number on the right or to the nearest left.
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Input
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the input line is an integer N (1 < n <= 100), giving the number of rows of the triangle. The following n lines give the number triangles. The number on the number triangle ranges between 0 and 100.
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Output
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outputs the maximum and.
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Sample input
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573 88 1 0 2 7 4 44 5 2 6 5
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Sample output
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30
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Source
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the questions translated from IOI 1994
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- Some summary:
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This problem was adopted in the
- Programming Guide. It adapts dynamic planning to solving the problem, which takes less space and time than the measure of recursion.
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- Although the algorithm was described clearly, I still got ' WA ' when first pubmitted my src.
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After
- I compared the standard answer with mine, I found then the standard one adds ten to the max boundary of the ARRA Y.
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Well
- , after checking, I find in the Circulation,i or J can be 101,so the max boundary must is no less than 101.
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The result always goes back to
- a careless
man ...
POJ 2760: Digital triangles