uva 562 - Dividing coins(01背包)
2013-09-04 13:43
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题目连接:562 - Dividing coins
题目大意:给出n个硬币的面值, 要求将这n个硬币分成两堆(不要求个数平均), 尽量使得两堆硬币的和之差的绝对值最小, 病输出最小值的绝对值。
解题思路:将sum 总和的一半half看成是背包的容量, 然后进行01背包处理, 然后取尽量接近half的可达到值cur, half - cur 的两倍就是所要求的值ans, 注意一点, 当sum 为奇数的时候, ans 要加 1 。
#include <stdio.h>
#include <string.h>
const int N = 50005;
const int M = 105;
int n, sum, dp
, coin[M], Max;
int solve() {
memset(dp, 0, sizeof(dp));
dp[0]= 1;
int half = sum / 2;
for (int i = 0; i < n; i++) {
for (int j = half - 1; j >= 0; j--)
if (dp[j]) dp[j + coin[i]] = 1;
}
for (int i = 0; i <= half; i++)
if (dp[half - i]) return i * 2;
}
int main() {
int cas;
scanf("%d", &cas);
while (cas--) {
sum = 0;
scanf("%d", &n);
memset(coin, 0, sizeof(coin));
for (int i = 0; i < n; i++) {
scanf("%d", &coin[i]);
sum += coin[i];
}
printf("%d\n", solve() + sum % 2);
}
return 0;
}
题目大意:给出n个硬币的面值, 要求将这n个硬币分成两堆(不要求个数平均), 尽量使得两堆硬币的和之差的绝对值最小, 病输出最小值的绝对值。
解题思路:将sum 总和的一半half看成是背包的容量, 然后进行01背包处理, 然后取尽量接近half的可达到值cur, half - cur 的两倍就是所要求的值ans, 注意一点, 当sum 为奇数的时候, ans 要加 1 。
#include <stdio.h>
#include <string.h>
const int N = 50005;
const int M = 105;
int n, sum, dp
, coin[M], Max;
int solve() {
memset(dp, 0, sizeof(dp));
dp[0]= 1;
int half = sum / 2;
for (int i = 0; i < n; i++) {
for (int j = half - 1; j >= 0; j--)
if (dp[j]) dp[j + coin[i]] = 1;
}
for (int i = 0; i <= half; i++)
if (dp[half - i]) return i * 2;
}
int main() {
int cas;
scanf("%d", &cas);
while (cas--) {
sum = 0;
scanf("%d", &n);
memset(coin, 0, sizeof(coin));
for (int i = 0; i < n; i++) {
scanf("%d", &coin[i]);
sum += coin[i];
}
printf("%d\n", solve() + sum % 2);
}
return 0;
}
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