Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16250 Accepted Submission(s): 5311
Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.
Given a consecutive number sequence S
1, S
2, S
3, S
4 ... S
x, ... S
n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S
x ≤ 32767). We define a function sum(i, j) = S
i + ... + S
j (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i
1, j
1) + sum(i
2, j
2) + sum(i
3, j
3) + ... + sum(i
m, j
m) maximal (i
x ≤ i
y ≤ j
x or i
x ≤ j
y ≤ j
x is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i
x, j
x)(1 ≤ x ≤ m) instead. ^_^
Input Each test case will begin with two integers m and n, followed by n integers S
1, S
2, S
3 ... S
n.
Process to the end of file.
Output Output the maximal summation described above in one line.
Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3
Sample Output
6
8
求区间的和。。DP题。。
#include
#include
#include
#include
#include
#include
#include
typedef __int64 LL; using namespace std; const int N=1000000+100; const int INF = 0x3f3f3f3f; LL a[N]; LL b[N]; LL dp[N]; LL Max; int main() { int m, n; while(scanf("%d%d", &m, &n)==2) { b[0] = dp[0] = 0; for(int i=1; i<=n; i++) { scanf("%I64d", a+i); b[i] = dp[i] = 0; } for(int i=1, j; i<=m; i++) { Max = -INF; for(j=i; j<=n; j++) { if(b[j-1]>dp[j-1]) dp[j] = b[j-1] + a[j]; else dp[j]=dp[j-1]+a[j]; b[j-1] = Max; if(dp[j]>Max) Max = dp[j]; } b[j-1] = Max; } printf("%I64d\n",Max); } return 0; }
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