Max Sum
Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence.
If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input
2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
Sample Output
Case 1: 14 1 4 Case 2: 7 1 6
就是让你求一个数组中，连续最大的子序列和，以及起始位置，终止位置。之前做过子序列和，就是用数组存前多少项的和，需要i~~j的子序列和，就用f[j]-f[i-1]就可以了。最大子序列和，只需得到前n项和的最大值，以及最小值就可以了。但是在维护的过程中发现，直接做，最小值可能出现在最大值之后，导致WA。修改了多次也没有AC，不得不另找方法。
思路：跟上面改变的不多，前n项和为负数的时候，就从第n+1项继续寻找最大值。代码如下：
#include<cstdio>
int main(){
int T,kase=1,n,x;
scanf("%d",&T);
while(T--){
int sum=0,maxsum=-9999,start=0,end=0,t=1;//start记录起点，end记录终点，t记录模拟起点
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&x);
sum+=x;
if(sum>maxsum){
maxsum=sum;
start=t;
end=i;
}
if(sum<0){//前i项为负数，从第i+1项重新做
sum=0;
t=i+1;
}
}
if(kase>1) printf("\n");
printf("Case %d:\n%d %d %d\n",kase++,maxsum,start,end);
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。