题目1442：A sequence of numbers

最新推荐文章于 2019-02-28 22:08:13 发布

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最新推荐文章于 2019-02-28 22:08:13 发布

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这篇博客讲述了关于帮助Xinlv解析旧纸上模糊的数列问题，内容涉及如何确定可能是等差或等比数列的前三项，并推算出序列中的其他数值。

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题目描述：

Xinlv wrote some sequences on the paper a long time ago, they might be arithmetic or geometric sequences. The numbers are not very clear now, and only the first three numbers of each sequence are recognizable. Xinlv wants to know some numbers in these sequences, and he needs your help.

输入：

The first line contains an integer N, indicting that there are N sequences. Each of the following N lines contain four integers. The first three indicating the first three numbers of the sequence, and the last one is K, indicating that we want to know the K-th numbers of the sequence.

You can assume 0 < K <= 10^9, and the other three numbers are in the range [0, 2^63). All the numbers of the sequences are integers. And the sequences are non-decreasing.

输出：

Output one line for each test case, that is, the K-th number module (%) 200907.

样例输入：

2
1 2 3 5
1 2 4 5

样例输出：

5
16

WA代码：

#include <stdio.h>
int main() {
    int n;
    while(scanf("%d",&n)!=EOF) {
        int i,k;
        long long x,y,z;
        for(i=0;i<n;i++) {
            scanf("%lld %lld %lld %d",&x,&y,&z,&k);
            bool arith=false,geo=false;
            int d,q;
            if((y-x)==(z-y)) {
                d = y - x;
                arith = true;
            }
            else {
                geo = true;
                q = y/x;
            }
            long long ans = 1;
            if(arith) 
                ans = x + (k-1)*d;
            if(geo) {
                k = k - 1;
                while(k) {
                    if(k%2==1) 
                        ans *= q;
                    q = q*q;
                    k /= 2;
                }
            }
            printf("%d\n",ans%200907);
        }
    }
    return 0;
}

错误原因：

1. 0 < K <= 10^9, and the other three numbers are in the range [0, 2^63)。 k可为int，x,y,z，ans要定义为long long，而且d,q也要定义为long long。

2.if(geo) {....} 中，q = q * q，不能直接计算出来，因为可能数会非常大，超过了long long的表示范围。

AC代码：

#include <stdio.h>
#define M 200907 // ans%M
int main() {
    int n;
    while(scanf("%d",&n)!=EOF) {
        int i,k;
        long long x,y,z;
        for(i=0;i<n;i++) {
            scanf("%lld %lld %lld %d",&x,&y,&z,&k);
            bool arith=false,geo=false;
            long long d,q;
            if((y-x)==(z-y)) {
                d = y - x;
                arith = true;
            }
            else {
                geo = true;
                q = y/x;
            }
            long long ans = x;
            if(arith) 
                ans = (x%M + ((k-1)%M*(d%M))%M)%M;
            if(geo) {
                k = k - 1;
                while(k) {
                    if(k%2==1) { 
                        ans = ans*q%M;
                    }
                    q = q*q % M;
                    k /= 2;
                }
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}

%符号的两个公式：

①(a * b) % c = (a % c) * (b % c)

②(a + b) % c = (a % c + b % c) % c