Problem B
Pay the Price
Input: standard input
Output: standard output
Time Limit: 2 seconds
Memory Limit: 32 MB
In ancient days there was a country whose people had very interesting habits. Some of them were lazy, some were very rich, some were very poor and some were miser. Obviously, some of the rich were miser (A poor was never miser as he had little to spend) and lazy but the poor were lazy as well (As the poor were lazy they remained poor forever). The following things were true for that country
a) As the rich were miser, no things price was more than 300 dollars (Yes! their currency was dollar).
b) As all people were lazy, the price of everything was integer (There were no cents and so beggars always earned at least one dollar)
c) The values of the coins were from 1 to 300 dollars, so that the rich (who were idle) could pay any price with a single coin.
Your job is to find out in how many ways one could pay a certain price using a limited number of coins (Note that the number of coins paid is limited but not the value or source. I mean there was infinite number of coins of all values). For example, by using three coins one can pay six dollars in 3 ways, 1+1+4, 1+2+3, and 2+2+2. Similarly, one can pay 6 dollars using 6 coins or less in 11 ways.
Input
The input file contains several lines of input. Each line of input may contain 1, 2 or 3 integers. The first integer is always N (0<=N<=300), the dollar amount to be paid. All other integers are less than 1001 and non-negative.
Output
For each line of input you should output a single integer.
When there is only one integer N as input, you should output in how many ways N dollars can be paid.
When there are two integers N and L1 as input, then you should output in how many ways N dollars can be paid using L1 or less coins.
When there are three integers N, L1 and L2 as input, then you should output in how many ways N dollars can be paid using L1, L1+1 …, L2coins (summing all together). Remember that L1 is not greater than L2.
Sample Input
6
6 3
6 2 5
6 1 6
Sample Output
11
7
9
11
题意：现在有1到300的硬币，有3种情况。首先输入n，表示要凑成的价值为n，然后如果后面没数字，求组成n的方式几种，如果后跟1个数字a，求用不多于a的硬币数组成n有几种，如果后跟2个数字a，b，求使用数量a到b的硬币组成方式有几种。
思路：凑硬币的变形，完全背包。dp数组多开一维用来表示用的硬币数。状态转移方程为：
dp[j][k] += dp[j - i][k - 1]。
代码：
#include <stdio.h>
#include <string.h>
int n, a, b, i, j, k;
long long dp[305][305], ans;
char sb[20];
int min(int a, int b) {
return a < b ? a : b;
}
int main() {
dp[0][0] = 1;
for (i = 1; i <= 300; i ++) {
for (j = i; j <= 300; j ++) {
for (k = 1; k <= 300; k ++) {
dp[j][k] += dp[j - i][k - 1];
}
}
}
while (gets(sb) != NULL) {
a = b = -1; ans = 0;
sscanf(sb, "%d%d%d", &n, &a, &b);
a = min(a, 300); b = min(b, 300);
if (a == -1) {
for (i = 0; i <= n; i ++) {
ans += dp[n][i];
}
}
else if (b == -1) {
for (i = 0; i <= a; i ++) {
ans += dp[n][i];
}
}
else {
for (i = a; i <= b; i ++) {
ans += dp[n][i];
}
}
printf("%lld
", ans);
}
return 0;
}