3.2常用技巧练习

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Links：https://vjudge.net/contest/147590#overview

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A - Sum of Consecutive Prime Numbers

求给定数字能表示成多少组质数之和，如47可以表示成2+3+5+7+11+13, 11+13+17, 41这三组。
先处理质数，然后尺取法。

#include<iostream>#include<string.h>#include<algorithm>#include<stdio.h>using namespace std;int prim[10000];int primnum;void isprim(){    prim[0] = 2;    primnum = 1;    for (int i = 3; i <= 10000; i++){        bool is = true;        for (int j = 0; j < primnum&&prim[j] < i; j++){            if (i%prim[j] == 0){                is = false; break;            }        }        if (is){            prim[primnum++] = i;        }    }}int main(){    isprim();    int n;    while (~scanf("%d", &n), n != 0){        int i = 0, j = 0;        int ans = 0;        int sum = prim[i];        while (prim[j] <= n){            if (sum == n){                ans++;                sum -= prim[i++];            }            else if(sum<n){                sum += prim[++j];            }            else if (sum>n){                sum -= prim[i++];            }        }        printf("%d\n", ans);    }    return 0;}

B - Graveyard Design

问给定数字能表示成多少组连续数字的平方和。
如：2030=25^2+26^2+27^2
=21^2+22^2+23^2+24^2
也是尺取法，一开始用数组存答案，不知为何会超时，后来看别人的改用stl容器和pair存答案就过了。

#include<iostream>#include<string.h>#include<algorithm>#include<vector>#include<stdio.h>using namespace std;vector<pair<long long, long long> > ans;//int ans[1000];//int ans2[1000];int ansnum;int main(){    long long n;    while (~scanf("%lld", &n)){        ans.clear();        long long i = 1, j = 1;        //ansnum = 0;        long long sum = i*i;        while (j*j <= n){            if (sum == n){                //ans[ansnum] = i;                //ans2[ansnum++] = j;                ans.push_back(make_pair(i, j));                j++;                sum += j*j;            }            else if (sum < n){                j++;                sum += j*j;            }            else {                sum -= i*i;                i++;            }        }        ansnum = ans.size();        printf("%d\n", ansnum);        for (int i = 0; i < ansnum; i++)        {            printf("%d", ans[i].second - ans[i].first+1);            for (int j = ans[i].first; j <= ans[i].second; j++)            {                printf(" %d", j);            }            puts("");        }    }    return 0;}

C - The Water Bowls

翻水杯，每次翻动会把左右两边的水杯翻转，为了不影响前面已经翻好的水杯，每当有一个水杯是1，就翻它后面那一个就行，分别从左边翻和右边翻，最小值为答案。

#include<iostream>#include<stdio.h>using namespace std;bool line[21];bool line2[21];int main(){    for (int i = 0; i < 20; i++){        cin >> line[i];        line2[i+1] = line[i];    }    int ans = 0;    int ans1 = 0;    for (int i = 0; i < 20; i++){        if (line[i] == 1){            line[i] =0;            line[i + 1] = (line[i + 1]==1?0:1);            line[i + 2] = (line[i + 2]==1?0:1);            ans++;        }    }    for (int i = 20; i >= 1; i--){        if (line2[i] == 1){            line2[i] = 0;            line2[i - 1] = line2[i - 1] ^ 1;            line2[i - 2] = line2[i - 2] ^ 1;            ans1++;        }    }    cout << (ans<ans1?ans:ans1);    return 0;}

D - EXTENDED LIGHTS OUT

二维版翻水杯，每次翻动会使上下左右四个都翻转，算法书上有讲过，枚举第一排的翻转方式，就可以确定下面的翻转，最后遍历看有没有全部翻转正确，得出答案。

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int map[7][8];int ans[7][8];int bmap[7][8];bool cal(int k){    memset(ans, 0, sizeof(ans));    for (int i = 1; i <= 5; i++){        for (int j = 1; j <= 6; j++){            bmap[i][j] = map[i][j];        }    }    for (int j = 1; j <= 6; j++){         if ((1<<j-1)&k){            ans[1][j] = 1;            bmap[1][j] ^= 1, bmap[1][j - 1] ^= 1, bmap[1][j + 1] ^= 1, bmap[2][j] ^= 1;        }    }    for (int i = 2; i <= 5; i++){        for (int j = 1; j <= 6; j++){            if (bmap[i - 1][j]){                ans[i][j] = 1;                bmap[i][j] ^= 1, bmap[i][j - 1] ^= 1, bmap[i][j + 1] ^= 1, bmap[i-1][j] ^= 1,bmap[i+1][j]^=1;            }        }    }    for (int i = 1; i <= 5; i++){        for (int j = 1; j <= 6; j++){            if (bmap[i][j]){ return false; }        }    }    return true;}int main(){    int n;    scanf("%d", &n);    for (int t = 0; t < n; t++){        for (int i = 1; i <= 5; i++){            for (int j = 1; j <= 6; j++){                scanf("%d", &map[i][j]);            }        }        for (int i = 0; i < 64;i++){            if (cal(i)){                break;            }        }        printf("PUZZLE #%d\n", t + 1);        for (int i = 1; i <= 5; i++){            for (int j = 1; j <= 6; j++){                printf("%d", ans[i][j]);                if (j == 6){ printf("\n"); }                else{ printf(" "); }            }        }    }    return 0;}

E - Sumsets

在给定数列里找出满足a+b+c=d。的最大的d
题解是变成a+b=d-c，满足c>a>b但是由于数字有正有负，所以d可能比abc都小。
所以枚举d和c，尺取法找a和b。
c从2开始枚举

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int num[1005];int main(){    int n;    while(~scanf("%d", &n),n!=0){        for (int i = 0; i < n; i++){            scanf("%d", &num[i]);        }        sort(num, num + n);        int sum,ans;        bool y = false;        for (int i = n - 1; i >=0; i--){            for (int j = 2; j <n; j++){                if (i == j){ continue; }                sum = num[i] - num[j];                int b = j - 1, a = 0;                while (a < b){                    if (sum == num[a] + num[b]){                        y = true;                        ans = i;                        break;                    }                    else if (sum > num[a] + num[b]){                        a++;                    }                    else{ b--; }                }                if (y == true)break;            }            if (y == true)break;        }        if (y == false){ printf("no solution\n"); }        else { printf("%d\n", num[ans]); }    }    return 0;}