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題目鏈接：　　　　https://codeforces.com/problemset/problem/1141/C

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C. Polycarp Restores Permutation time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output An array of integers p1,p2,…,pnp1,p2,…,pn is called a permutation if it contains each number from 11 to nn exactly once. For example, the following arrays are permutations: [3,1,2][3,1,2], [1][1], [1,2,3,4,5][1,2,3,4,5] and [4,3,1,2][4,3,1,2]. The following arrays are not permutations: [2][2], [1,1][1,1], [2,3,4][2,3,4].Polycarp invented a really cool permutation p1,p2,…,pnp1,p2,…,pn of length nn. It is very disappointing, but he forgot this permutation. He only remembers the array q1,q2,…,qn?1q1,q2,…,qn?1 of length n?1n?1, where qi=pi+1?piqi=pi+1?pi.Given nn and q=q1,q2,…,qn?1q=q1,q2,…,qn?1, help Polycarp restore the invented permutation.Input The first line contains the integer nn (2≤n≤2?1052≤n≤2?105) — the length of the permutation to restore. The second line contains n?1n?1 integers q1,q2,…,qn?1q1,q2,…,qn?1 (?n<qi<n?n<qi<n).Output Print the integer -1 if there is no such permutation of length nn which corresponds to the given array qq. Otherwise, if it exists, print p1,p2,…,pnp1,p2,…,pn. Print any such permutation if there are many of them.
Examples

input 3 -2 1 output 3 1 2

input 5 1 1 1 1 output 1 2 3 4 5

input 4 -1 2 2 output -1

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題意：? ? ?輸入整數(shù)n? ，在輸入n-1個(gè)數(shù)，代表著目標(biāo)數(shù)組后一個(gè)減前一個(gè)數(shù)組的差的值，　　即：　ans[i+1] - ans[i]

　　　　若存在則輸出數(shù)組ans的值　　　　　　若不存在則輸出-1

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思路： 　暴力水題。。。。。。。。。。。。。。

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#include<iostream> #include<cstdio> #include<ctime> #include<cstring> #include<cstdlib> #include<cmath> #include<queue> #include<stack> #include<map> #include<algorithm> #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) #define Mem0(x) memset(x,0,sizeof(x)) #define Mem1(x) memset(x,-1,sizeof(x)) #define MemX(x) memset(x,0x3f,sizeof(x)) using namespace std; typedef long long ll; const int inf=0x3f3f3f; const double pi=acos(-1.0);ll ans[2000010],q[2000010],cnt[2000010],n; int main() {memset(cnt,0,sizeof(cnt));scanf("%lld",&n);ll tmp=0,l=0;for (int i=1;i<n;i++){scanf("%lld",&q[i]);l+=q[i]; tmp=tmp+l;}ll temp=(n+1)*n/2;ll pp=temp-tmp;bool flag=false;temp=pp/n;for (int i=1;i<=n;i++){ans[i]=temp;if (temp<=0||temp>n||cnt[ans[i]]>=1){flag=true;}else{cnt[ans[i]]++;}temp=temp+q[i];}if (flag)printf("-1");else{for (int i=1;i<=n;i++){printf("%lld ",ans[i]);}}printf("\n");return 0; }

　　

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轉(zhuǎn)載于:https://www.cnblogs.com/q1204675546/p/10951678.html

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