HDU 1267 下沙的沙子有几粒?

题目链接:下沙的沙子有几粒?

解题思路:这题和1133的差不多,只不过每一个H和D都没有分别,所以上一题的基础之上,除去n!和m!。

#include<stdio.h>
#include<string.h>
#define MAX 1000

typedef struct A{
    int num[MAX];
    int len;
    A(){
        memset(num, 0, sizeof(num));
        len = 0;
    }
    A(int a){
        memset(num, 0, sizeof(num));
        len = 0;
        while(a){
            num[len++] = a % 10;
            a /= 10;
        }
    }
    A(char kk[]){
        int i, j, k;
        memset(num, 0, sizeof(num));
        len = 0;
        for(i = strlen(kk) - 1; i >= 0; i--){
            num[len++] = kk[i] - '0';
        }
    }
}Big_int;

Big_int add(Big_int a, Big_int b){
    Big_int ret = Big_int();
    int rem = 0;
    int i, j, k;
    for(i = 0; i < MAX; i++){
        if(!rem && !a.num[i] && !b.num[i]){
            ret.len = i;
            break;
        }
        ret.num[i] = (rem + a.num[i] + b.num[i]) % 10;
        rem = (rem + a.num[i] + b.num[i]) / 10;
    }
    return ret;
}

Big_int subtract(Big_int a, Big_int b){
    Big_int ret = Big_int();
    int i, j, k;
    for(i = 0; i < MAX; i++){
        if(!a.num[i] && !b.num[i]){
            ret.len = i;
            break; 
        }
        if(a.num[i] < b.num[i]){
            int tem = i + 1;
            while(!a.num[tem]){
                a.num[tem] = 9;
                tem++;
            }
            a.num[tem]--;
            ret.num[i] = 10 + a.num[i] - b.num[i];
        }
        else{
            ret.num[i] = a.num[i] - b.num[i];
        }
    }
    return ret;
}

Big_int multiply(Big_int a, Big_int b){
    Big_int ret = Big_int();
    int i, j, k;
    for(i = 0; i < a.len; i++){
        for(j = 0; j < b.len; j++){
            ret.num[i + j] += a.num[i] * b.num[j];
        }
    }
    int rem = 0;
    for(i = 0; i < MAX; i++){
        int tem = rem + ret.num[i];
        ret.num[i] = tem % 10;
        rem = tem / 10;
    }
    for(i = MAX - 1; ; i--){
        if(ret.num[i] != 0){
            ret.len = i + 1;
            break;
        }
    }
    return ret;
}

Big_int divide(Big_int a, int b){
    Big_int ret;
    int i, j, k;
    int rem = 0;
    for(i = a.len - 1; i >= 0; i--){
        ret.num[i] = (rem * 10 + a.num[i]) / b;
        rem = (rem * 10 + a.num[i]) % b;
    }
    for(i = MAX - 1; ; i--){
        if(ret.num[i] != 0){
            ret.len = i + 1;
            break;
        }
    }
    return ret;
}

int mod(Big_int a, int b){
    int i, j, k, ret = 0;
    for(i = a.len - 1; i >= 0; i--){
        ret = (a.num[i] % b + ((ret % b) * (10 % b)) % b) % b;    
    }
    return ret;
}

Big_int power(Big_int a, int b){
    Big_int ret = Big_int(1);
    int i, j, k;
    Big_int tem = Big_int(1);
    tem = multiply(tem, a);
    while(b){
        if(b & 1){
            ret = multiply(ret, tem);
        }
        b >>= 1;
        tem = multiply(tem, tem);
    }
    for(i = MAX - 1; ; i--){
        if(ret.num[i] != 0){
            ret.len = i + 1;
            break;
        }
    }
    return ret;
}

void display(Big_int a){
    int i, j, k;
    if(a.len == 0){
        printf("0");
    }
    else{
        for(i = a.len - 1; i >= 0; i--){
            printf("%d", a.num[i]);
        }    
    }    
    printf("\n");
}


int main(){
    int i, j, k;
    int m, n, tot = 1;
    while(scanf("%d%d", &m, &n) != EOF){
        Big_int ans = Big_int(1);
        Big_int tem1, tem2;
        for(i = 1; i <= m + n; i++){
            ans = multiply(ans, Big_int(i));
        }
        ans = multiply(ans, Big_int(m + 1 - n));
        //display(ans);
        for(i = 2; i <= m + 1; i++){
        	if(i > n){
	        	ans = divide(ans, i);
				continue;	
	        } 
        	ans = divide(ans, i * i);
        	
        }
        display(ans);
    }
    return 0;
}


 

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