uva10759 手算多项式分解 或者dp

#include <iostream>#include <cstdio>#include <cstring>using namespace std;typedef unsigned long long ULL;ULL a[555];ULL b[555];ULL gcd(ULL x, ULL y){    return y?gcd(y,x%y):x;}/**手算多项式分解把原题变为(x^1+x^2+x^3+x^4+x^5+x^6)^n 求指数大于等于x的项数的系数之和**/int main(){//    freopen("data.in", "r", stdin);    int n, x;    ULL ans1, ans2;    while(scanf("%d%d", &n, &x) != EOF && n+x){        memset(a, 0 ,sizeof a);        for(int i = 1; i <= 6; i++)            a[i] = 1;        for(int k = 1; k < n; k++){            memset(b, 0, sizeof b);            for(int i = k; i <= 6*k; i++){                for(int j = 1; j <= 6; j++){                    b[i+j] += a[i];                }            }            memcpy(a, b, sizeof b);        }        ans1 = ans2 = 0;        for(int i = 0; i <= n*6; i++){            if(i >= x) ans1 += a[i];            ans2 += a[i];        }        if(ans1 == 0)puts("0");        else if(ans1 == ans2)puts("1");//        else printf("%llu/%llu\n", ans1, ans2);        else printf("%llu/%llu\n", ans1/gcd(ans1, ans2), ans2/gcd(ans1, ans2));    }    return 0;}

dp的做法是用f[i][j]存投掷i次得分为j的概率 转移方程很简单  重点是用分数表示 记得用LCM GCD处理同分和约分  保证分数加减的正确性

#include <iostream>#include <cstdio>#include <cstring>using namespace std;typedef unsigned long long ULL;ULL f[25][155], e[25][155];ULL sum1[25][155], sum2[25][155];ULL gcd(ULL x, ULL y){    return y?gcd(y,x%y):x;}ULL lcm(ULL x, ULL y){    return x/gcd(x,y)*y;}int main(){//    freopen("data.in", "r", stdin);    f[0][0] = e[0][0] = sum1[0][0] = sum2[0][0] = 1;    for(int i = 1; i <= 24; i++){        for(int j = i; j <= 6*i; j++){            ULL FZ = 0, FM = 1, TMP;            for(int k = max(0, j-6); k < j; k++){                if(f[i-1][k] == 0) continue;                TMP = lcm(FM, e[i-1][k]*6);                FZ = FZ*(TMP/FM) + f[i-1][k]*(TMP/(e[i-1][k]*6));                FM = TMP;                TMP = gcd(FZ, FM);                FZ /= TMP, FM /= TMP;            }            f[i][j] = FZ, e[i][j] = FM;            if(i == j){                sum1[i][j] = f[i][j];                sum2[i][j] = e[i][j];                continue;            }            TMP = lcm(sum2[i][j-1], e[i][j]);            FZ = sum1[i][j-1]*(TMP/sum2[i][j-1])+f[i][j]*(TMP/e[i][j]);            FM = TMP;            TMP = gcd(FZ, FM);            sum1[i][j] = FZ/TMP, sum2[i][j] = FM/TMP;        }    }    int n, x;    ULL ans1, ans2;    while(scanf("%d%d", &n, &x) != EOF && n+x){        if(x > 6*n) puts("0");        else if(x <= n) puts("1");        else{            ans1 = sum2[n][x-1] - sum1[n][x-1];            ans2 = sum2[n][x-1];            printf("%llu/%llu\n", ans1, ans2);        }    }    return 0;}