最大連續欄位和
阿新 • • 發佈:2018-11-18
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.#include<cstdio> #include<algorithm> #include<cstring> #include<iostream> #include<cmath> #include<map> #include<stack> #include<vector> #include<queue> #include<set> using namespace std; const int MAX_N=100005; #define inf 1<<23 typedef long long ll; typedef long long LL; int a[MAX_N]; int n; int s,e;//記錄開始和結束 void text() { int maxValue=0; int thisMaxValue=0; for(int i=0;i<n;i++) { thisMaxValue+=a[i]; if(thisMaxValue>0) { maxValue=thisMaxValue>maxValue?thisMaxValue:maxValue; } else { thisMaxValue=0; } } printf("%d\n",maxValue); } int main() { while(scanf("%d",&n)&&n) { for(int i=0;i<n;i++) { scanf("%d",&a[i]); } text(); } return 0; }
InputThe 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).
OutputFor 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
Case 1: 14 1 4 Case 2: 7 1 6
#include<cstdio> #include<algorithm> #include<cstring> #include<iostream> #include<cmath> #include<map> #include<stack> #include<vector> #include<queue> using namespace std; const int MAX_N=100010; typedef long long ll; typedef long long LL; int a[MAX_N]; int n; void text(int &res,int &s,int &e) { int maxValue=-(1<<30); int thisMaxValue=0; int k=1; for(int i=0; i<n; i++) { thisMaxValue+=a[i]; //cout<< thisMaxValue <<endl; if(thisMaxValue>maxValue) { maxValue=thisMaxValue; s=k; e=i+1; } if(thisMaxValue<0) { thisMaxValue=0; k=i+2; } } res=maxValue; } int main() { int t; int s,e; int res; scanf("%d",&t); for(int j=1; j<=t; j++) { s=0; e=0; memset(a,0,sizeof(a)); scanf("%d",&n); for(int i=0; i<n; i++) { scanf("%d",&a[i]); } text(res,s,e); if(j!=t) { printf("Case %d:\n",j); printf("%d %d %d\n\n",res,s,e); } else { printf("Case %d:\n",j); printf("%d %d %d\n",res,s,e); } } return 0; }