HDU1231：
给定K个整数的序列{ N1, N2, …, NK }，其任意连续子序列可表示为{ Ni, Ni+1, …, Nj }，其中 1 <= i <= j <= K。最大连续子序列是所有连续子序列中元素和最大的一个， 例如给定序列{ -2, 11, -4, 13, -5, -2 }，其最大连续子序列为{ 11, -4, 13 }，最大和 为20。 在今年的数据结构考卷中，要求编写程序得到最大和，现在增加一个要求，即还需要输出该 子序列的第一个和最后一个元素。 Input 测试输入包含若干测试用例，每个测试用例占2行，第1行给出正整数K( < 10000 )，第2行给出K个整数，中间用空格分隔。当K为0时，输入结束，该用例不被处理。 Output 对每个测试用例，在1行里输出最大和、最大连续子序列的第一个和最后一个元 素，中间用空格分隔。如果最大连续子序列不唯一，则输出序号i和j最小的那个（如输入样例的第2、3组）。若所有K个元素都是负数，则定义其最大和为0，输出整个序列的首尾元素。 Sample Input 6 -2 11 -4 13 -5 -2 10 -10 1 2 3 4 -5 -23 3 7 -21 6 5 -8 3 2 5 0 1 10 3 -1 -5 -2 3 -1 0 -2 0 Sample Output 20 11 13 10 1 4 10 3 5 10 10 10 0 -1 -2 0 0 0
Huge input, scanf is recommended. Hint Hint
题意：给出一行数字，找出一串和最大的连续的子序列。
解题思路：模拟心中的想法。当当前子序列的和小于0时这串子序列就不再继续往下扩展了，此时子序列的开头应从下一个元素开始了。如5 -8 3 2 5 0，当以5为序列开头，计算到5-8=-3<0时就不再继续扩招将3加入子序列了（想一想为什么）而是将3作为子序列新的开头往后判断。
#include<stdio.h> int main() { int a[10010],t,sum,max,left,right,i,n; while(~scanf("%d",&n)&&n) { for(i=0;i<n;i++) scanf("%d",&a[i]); t=a[0]; left=right=a[0]; sum=0; max=a[0]; for(i=0;i<n;i++) { if(sum<0) { t=a[i]; sum=a[i]; } else { sum+=a[i]; } if(sum>max) { max=sum; left=t; right=a[i]; } } if(max<0) printf("0 %d %d\n",a[0],a[n-1]); else printf("%d %d %d\n",max,left,right); } return 0; }Uva108：
A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem. Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the subrectangle with the largest sum is referred to as the maximal sub-rectangle. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array. As an example, the maximal sub-rectangle of the array: 0 −2 −7 0 9 2 −6 2 −4 1 −4 1 −1 8 0 −2 is in the lower-left-hand corner: 9 2 −4 1 −1 8 and has the sum of 15. Input The input consists of an N × N array of integers. The input begins with a single positive integer N on a line by itself indicating the size of the square two dimensional array. This is followed by N2 integers separated by white-space (newlines and spaces). These N2 integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on the second row, left-to-right, etc.). N may be as large as 100. The numbers in the array will be in the range [−127, 127]. Output The output is the sum of the maximal sub-rectangle. Sample Input 4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2 Sample Output 15
题意：给出一个矩阵，找出矩阵中和最大的子矩阵。
解题思路:着重思考一下这道题。受上道题启发，可不可以将矩阵也用处理序列的方法来解决这道题？ 举个例子：
（图虽然有点粗糙但是应该可以看得懂~）
将输入的数字进行以下处理：a[i][j]+=a[i-1][j]; 使a数组中每个数表示到该行为止该列的和，如图中5对应0+9+（-4）
通过三个for循环使得每一块子矩阵都可以被表示。
再使用上题中的思路模拟求解。
#include<stdio.h> int main() { int a[110][110],n,t,i,j,k,max,sum; while(~scanf("%d",&n)) { for(i=1;i<=n;i++) for(j=1;j<=n;j++) { scanf("%d",&a[i][j]); a[i][j]+=a[i-1][j]; } max=a[1][1]; for(i=0;i<=n;i++) { for(j=i;j<=n;j++) { sum=0; for(k=1;k<=n;k++) { if(sum<0) sum=0; else if(i!=j) sum+=a[j][k]-a[i][k]; if(sum>max&&sum!=0) max=sum; } } } printf("%d\n",max); } return 0; }