pat_1007

来源：互联网发布：软件远程升级方案编辑：程序博客网时间：2024/06/05 12:59

源地址：http://www.patest.cn/contests/pat-a-practise/1007

最大子段和。。刚刚打过模板。。所以记得很清楚。。不过刚开始wa了一发，是因为题目特殊定义说如果这个数列全为负数，那么定义成最大子段和为0，且输出首尾的数字，在这有一个判断条件写的不对。。具体看代码。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<set>#include<map>//#define ONLINE_JUDGE#define eps 1e-3#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m;#define Mod 1000000007#define N 510#define M 1000100const int size = 10010;const int mod = 9901;int a[10010];int p[10010];void maxSub(){int pos=0,mmax=0;memset(p,0,sizeof p);int si_1,si;si_1=mmax=a[0];p[0] = 0;for(int i=1;i<n;i++){if(si_1<0){si = a[i];p[i] = 0;}else{si = si_1+a[i];p[i] = 1;}si_1 = si;if(si>mmax){mmax = si;pos = i;}}int k;for(k=pos;k>=0;k--)if(p[k]==0) break;printf("%d %d %d\n",mmax,a[k],a[pos]);}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif    while(scanf("%d",&n)!=EOF){    int flag=0;    for(int i=0;i<n;i++){    scanf("%d",&a[i]);    if(a[i]>=0)//>= instead of >    flag = 1;    }    if(!flag){    printf("0 %d %d\n",a[0],a[n-1]);    continue;    }    maxSub();    }return 0;}

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