poj 3181 Dollar Dayz

题目大意:

N由 1到K的数组成，求这样的组成方式有多少种？

Input

A single line with two space-separated integers: N and K.

Output

A single line with a single integer that is the number of unique ways FJ can spend his money.

Sample Input

5 3

Sample Output

5

用动态规划来解此题
dp[i][j]表示前i个数构成j的方法数
若j < i
则dp[i][j] = dp[i-1][j]
否则
dp[i][j] = dp[i][j-i] + dp[i-1][j]

还可以简化为一维数组
代码如下

#include <cstdio>
#include <cstring>

int dp[1002];

int main(int argc, char const *argv[])
{
    int n, k;
    scanf("%d %d",&n,&k);
    memset(dp, 0, sizeof(dp));
    dp[0] = 1;
    for(int i = 1; i <= k; i++) {
        for(int j = i; j <= n; j++) {
            dp[j] = dp[j] + dp[j-i];
        }
    }
    printf("%d\n",dp[n]);
    return 0;
}

但提交错误，考虑到数值太大溢出，只好模拟大整数的加法

设一位代表5位

那么int dp[1002][20]每一个数可以最多有100位，代码如下

#include <cstdio>
#include <cstring>
#define base 100000

int dp[1002][20];

void add(int a, int b) {
    int ci = 0;
    for(int i = 0; i < 20; i++) {
        int ben = dp[a][i] + dp[b][i] + ci;
        dp[a][i] = ben % base;
        ci = ben/base;
    }
}

void show(int n) {
    bool isF = true;
    for(int i = 19; i >= 0; i--) {
        if(isF && dp[n][i] != 0) {
            isF = false;
            printf("%d",dp[n][i]);//第一个不用凑够5位
        }
        else if(!isF) {
            printf("%05d",dp[n][i]);//注意要凑够5位，不够前面补0
        }
    }
    puts("");
}
int main(int argc, char const *argv[])
{
    int n, k;
    scanf("%d %d",&n,&k);
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for(int i = 1; i <= k; i++) {
            for(int j = i; j <= n; j++) {
                add(j, j-i);
            }
        }
        show(n);
    
    
    return 0;
}