NOIP 2005 过河 解题报告

　　这题考的是压缩，当两个石子之间的距离大于100的时候就让他们变成100，然后计算就十分方便了，代码如下：

#include <stdio.h>
#include <stdlib.h>
#define min(a, b) ((a)<(b)?(a):(b))
int num[101];
int f[10101];
int stone[10101];

int com(const void *a, const void *b)
{
	return *(int *)a - *(int *)b;
}

int main(int argc, char **argv)
{
	int i, j, k;
	int l, s, t, m;
	scanf("%d%d%d%d", &l, &s, &t, &m);
	for(i = 1; i <= m; i++){
		scanf("%d", &num[i]);
	}
	if(s == t){
		for(i = 1, j = 0; i <= m; i++){
			if(num[i] % s == 0){
				j++;
			}
		}
		printf("%d\n", j);
		return 0;
	}
	qsort(num, m + 1, sizeof(int), com);
	for(i = 1, j = 0; i <= m; i++){
		if(num[i] - num[i - 1] > 100){
			j += 100;
			stone[j] = 1;
		}else{
			j += num[i] - num[i - 1];
			stone[j] = 1;
		}
	}
	k = j + 100;
	for(i = 1; i <= k; i++){
		f[i] = 0xFFFFFFF;
		for(j = s; j <= t; j++){
			if(i - j < 0){
				break;
			}
			f[i] = min(f[i], f[i - j] + stone[i]);
		}
	}
	printf("%d\n", f[k]);
	return 0;
}