Nine degree topic 1453:greedy Tino

Title Description:

Tino wrote a long long story. but! In Chinese ...
So I has the to tell you the problem directly and discard he long long story. That's Tino want to carry some oranges with "carrying pole", and he must make-side of the carrying pole are the same Weight. Each orange has its ' weight. So greedy Tino want to know the maximum weight he can carry.

input:

the first line of input contains a number t, which means there is t cases of the test data.
The first line contain a number n, indicate the number of oranges.
The second line contains n numbers, Wi, indicate the weight of all orange
N is between 1 and inclusive. Wi is between 0 and inclusive. The sum of Wi is equal or less than.

Output:

For each test case, output the maximum weight in one side of carrying pole. If you can ' t carry any orange, output-1. Output format is shown in Sample output.

Sample input:

151 2 3) 4 5

Sample output:

Case 1:7

The dynamic programming algorithm solves the transfer equation as follows:

The equation means: Select in the first I items, divided into two piles, the first heap than the second pile of heavy J kg, the maximum total weight of the two piles combined! (Note: J can be negative)

The code is as follows:

#include <stdio.h> #include <string.h> #define OFFSET 2000//because J has positive or negative, so add an offset to make the title # define INF 0x7fffffffint Wi[101];int dp[101][4001];int Max (int A, int b, int c) {    int flag = a;    if (flag < b) {  &NB Sp     flag = b;   }    if (flag < C) {        flag = c;   }&N Bsp   return flag;} void Init () {    for (int i=0;i<40001;i++)             Dp[0][i] =-inf;  &nbs P Dp[0][offset] = 0;//core}int main () {    int T, n;    scanf ("%d", &t);    for (int i = 0; I < T; i++) {        scanf ("%d", &n);        wi[0] = 0;        &NB Sp;bool Haszero = false;        int cnt = 0;        for (int j = 0; J < N; j + +) {             scanf ("%d", &wi[++cnt]);          &NBSP if (wi[cnt] = = 0) {                   haszero = true;                   cnt--;           }      &NBS P  }        n = cnt;        init ();        for (int j = 1; J <= N J + +) {            for (int k = -2000; k <=; k++) {          &N Bsp     Dp[j][k+offset] = max (Dp[j-1][k-wi[j]+offset]+wi[j], dp[j-1][k+wi[j]+offset]+wi[j], Dp[j-1][k+offset]) ;           }       }        printf ("Case%d:", i+1);        if (Dp[n][0+offset] = 0) {            if (Haszero = True) {&N Bsp               printf ("0\n");           }else{  &nbs P            printf (" -1\n");           }       } else{            printf ("%d\n", DP[N][0+OFFSET]/2);       }  &nbsp ; }    return 0;}

Nine degree topic 1453:greedy Tino