1007. Maximum Subsequence Sum (25)
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.
Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.
Input Specification:
Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.
Output Specification:
For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.
Sample Input:10 -10 1 2 3 4 -5 -23 3 7 -21Sample Output:
10 1 4
方法一：
1 #include <cstdio> 2 #include <cstring> 3 #include <string> 4 #include <queue> 5 #include <stack> 6 #include <iostream> 7 using namespace std; 8 #define size 10000 9 int line[size+5]; 10 int main(){ 11 //freopen("D:\INPUT.txt","r",stdin); 12 int k; 13 int i,j,l,fl,fr; 14 bool t=false; 15 scanf("%d",&k); 16 for(i=0;i<k;i++){ 17 scanf("%d",&line[i]); 18 if(line[i]>=0){ 19 t=true; 20 } 21 } 22 /*if(!t){ 23 cout<<0<<" "<<line[0]<<" "<<line[k-1]<<endl; 24 return 0; 25 }*/ 26 long long tempmax=line[0],fmax=line[0]; 27 l=0; 28 fl=fr=0; //这句话没加 case3（从0开始）过不了 29 line[k]=0; 30 for(i=1;i<=k;i++){ 31 if(tempmax>=0){ 32 if(tempmax>fmax){ 33 fmax=tempmax; 34 fl=l; 35 fr=i-1; 36 } 37 }else{ 38 tempmax=0; 39 l=i; 40 } 41 tempmax+=line[i]; 42 } 43 if(fmax<0){ 44 fmax=0; 45 fl=0; 46 fr=k-1; 47 } 48 cout<<fmax<<" "<<line[fl]<<" "<<line[fr]<<endl; 49 return 0; 50 }
方法二要比方法三简明，更容易理解。
方法二：
1 #include <stdio.h> 2 #include <string.h> 3 int num[10005]; 4 int main(){ 5 //freopen("D:\INPUT.txt","r",stdin); 6 int n; 7 scanf("%d",&n); 8 int i,j=0; 9 for(i=0;i<n;i++){ 10 scanf("%d",&num[i]); 11 if(num[i]<0){ 12 j++; 13 } 14 } 15 if(j==n){ 16 printf("%d %d %d ",0,num[0],num[n-1]); 17 } 18 else{ 19 int tempsum=0; 20 int tempi=0,tempj; 21 int sum,l,r; 22 sum=l=0; 23 r=n-1; 24 for(i=0;i<n;i++){ 25 if(tempsum>=0){ 26 tempsum+=num[tempj=i]; 27 }else{ 28 tempsum=num[tempi=tempj=i]; 29 } 30 if(tempsum>sum||(tempsum==0&&r==n-1)){//注意单个0的情况 31 sum=tempsum; 32 l=tempi; 33 r=tempj; 34 } 35 } 36 printf("%d %d %d ",sum,num[l],num[r]); 37 } 38 return 0; 39 }
方法三：
1 #include <stdio.h> 2 #include <string.h> 3 int num[10005]; 4 int main(){ 5 //freopen("D:\INPUT.txt","r",stdin); 6 int n; 7 scanf("%d",&n); 8 int i,j=0; 9 for(i=0;i<n;i++){ 10 scanf("%d",&num[i]); 11 if(num[i]<0){ 12 j++; 13 } 14 } 15 if(j==n){ 16 printf("%d %d %d ",0,num[0],num[n-1]); 17 } 18 else{ 19 num[n]=0; 20 int tempsum=0; 21 int tempi,tempj; 22 int sum,l,r; 23 sum=l=0; 24 r=n-1; 25 for(i=0;i<=n;i++){ 26 if(tempsum>0){ 27 if(tempsum>sum){//当i元素前面的局部和>0时，有可能更新 28 sum=tempsum; 29 l=tempi; 30 r=tempj; 31 } 32 tempsum+=num[tempj=i]; 33 }else{//否则，i元素前面的局部和<=0 34 if(tempsum==0&&sum==0&&tempi==tempj&&l!=r){//i元素前面的局部和==0，并且局部和只有一个元素，并且最终和未更新，则更新最新和 35 l=r=tempi; 36 tempsum+=num[tempj=i]; 37 } 38 else{//其他情况，一律不更新 39 tempi=tempj=i; 40 tempsum=num[i]; 41 } 42 } 43 } 44 printf("%d %d %d ",sum,num[l],num[r]); 45 } 46 return 0; 47 }