poj 1850 Code

一个组合问题。

现求出比所给字符串str短或和它一样长的所有结果，然后把和str一样长并且比str大的减去，求比str大的字符串时，从i = 1到len - 1比较，只要当前位比str的当前位大即可。

/*
* Author: stormdpzh
* Time: 2012/5/10 16:53:48
* POJ: 1850 Code
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <functional>

#define sz(v) ((int)(v).size())
#define rep(i, n) for(int i = 0; i < n; i++)
#define repf(i, a, b) for(int i = a; i <= b; i++)
#define repd(i, a, b) for(int i = a; i >= b; i--)
#define out(n) printf("%d\n", n)
#define wh(n) while(scanf("%d", &n) != EOF)
#define whz(n) while(scanf("%d", &n) != EOF && n != 0)
#define lint long long

using namespace std;

char str[15];
int tri[30][30];
int res;

void getTri()
{
memset(tri, 0, sizeof(tri));
tri[0][0] = 1;
repf(i, 1, 29) rep(j, i + 1) {
if(j == 0 || j == i)
tri[i][j] = 1;
else
tri[i][j] = tri[i - 1][j - 1] + tri[i - 1][j];
}
}

int main()
{
getTri();
while(scanf("%s", str) != EOF) {
res = 0;
int len = strlen(str);
rep(i, len) repf(j, i + 1, len - 1) {
if(str[i] > str[j]) {
printf("0\n");
return 0;
}
}

repf(i, 1, len)
res += tri[26][i];
rep(i, len)
res -= tri[26 - (str[i] - 'a' + 1)][len - i];
out(res);
}

return 0;
}