Max Sum Plus Plus
Time
Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7919 Accepted Submission(s): 2624
Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (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(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Process to the end of file.
#include <stdio.h>
int f[1000003];
int best[2][1000003];
int N, M; //m組 n個數字
int a[1000003];
int sum[1000003];
int max(int,int);
int main() {
while (scanf("%d %d", &M, &N) != EOF) {
int ans = -2000000000;
int i,j;
sum[0] = 0;
for (i = 1; i <= N; i++) {
scanf("%d", &a[i]);
sum[i] = sum[i - 1] + a[i]; //sum[i]式從0到i a[0] + ...a[i]
}
for (i = 0; i <= N; i++) best[0][i] = 0;
int cur = 0;
for (j = 1; j <= M; j++) {
for (i = 0; i <= j - 1; i++) best[1 - cur][i] = -2000000000;
for (i = j; i <= N; i++) {
if (j == i) f[i] = sum[i]; //i是數組,j式分段數,如果i==j那麼一定是總數的和
else f[i] = max( best[cur][i - 1], f[i - 1] ) + a[i];//f[i]表示在分j段、i個數的情況下,最大值。 best[cur][i-1] 表示j-1段 i-1個數的最大值,然後把a[i]單獨當成第j段,f[-1]表示j段 i-1個數的最大值,加上a[i]表示沒有把a[i]
//當成第j段,而是第j段中的一個
best[1 - cur][i] = max(best[1 - cur][i - 1], f[i]);//best[1-cur][i]表示 分成j段 i個數的最大值
if (j == M) ans = max(ans, f[i]);
}
cur = 1 - cur;
}
printf("%d\n", ans);
}
return 0;
}
int max(int a,int b)
{
return (a>b)?a:b;
}
scanf("%d", &a[i]);
sum[i] = sum[i - 1] + a[i]; //sum[i]式從0到i a[0] + ...a[i]
}
for (i = 0; i <= j - 1; i++) best[1 - cur][i] = -2000000000;
for (i = j; i <= N; i++) {
if (j == i) f[i] = sum[i];
else f[i] = max( best[cur][i - 1], f[i - 1] ) + a[i];
best[1 - cur][i] = max(best[1 - cur][i - 1], f[i]);
if (j == M) ans = max(ans, f[i]);
}
cur = 1 - cur;
}